List containing indexes of equal elements - list

Question

I'm trying to define a predicate that takes 3 lists as parameters and the third list contains the indexes where the elements in the other two lists are the same.
This is my attempt at solving the problem.
sameIndex(List1, List2, List3) :- findIndex(List1,List2,0, List3).
% Helper Function
% Base cases
findIndex([],_,_,[]).
findIndex(_,[],_,[]).
findIndex([Head1|Tail1], [Head2|Tail2], Count, List4):-
Head1 == Head2, append([Count], List4, FinalList), NewCount is Count+1,
findIndex(Tail1,Tail2,NewCount,FinalList); NewCount is Count+1,
findIndex(Tail1,Tail2,NewCount, List4).
Sample test case:
sameIndex([1,2,3,4], [6,2,4,4], List)
should return
List = [1,3]
My logic is : if the heads of the lists are equal then append Count (keeps track of the index we're at) to our empty List4, increment Count, and recursively call the predicate with the tails of the two lists. Otherwise, increment Count and recursively call the predicate with the tails.
I'm assuming my code has improper use of arithmetic in prolog but I just can't get it to work. Any suggestions/help is appreciated.

Answer 1

Here is your code corrected - and reformatted, using 'standard' indentation
findIndex([Head1|Tail1], [Head2|Tail2], Count, List4):-
( Head1 == Head2,
append([Count], FinalList, List4),
NewCount is Count+1,
findIndex(Tail1,Tail2,NewCount,FinalList)
; NewCount is Count+1,
findIndex(Tail1,Tail2,NewCount, List4)
).
Your error was to invert the arguments to append/3.
Another problem it's returning multiple solutions - some of them wrong, so you should use the if/then/else construct, and discriminate the 2 base cases:
findIndex([],_,_,[]) :- !.
findIndex(_,[],_,[]).
findIndex([Head1|Tail1], [Head2|Tail2], Count, List4):-
( Head1 == Head2
->append([Count], FinalList, List4),
NewCount is Count+1,
findIndex(Tail1,Tail2,NewCount,FinalList)
; NewCount is Count+1,
findIndex(Tail1,Tail2,NewCount, List4)
).
But the code can be far simpler, using some library helpers:
sameIndex(List1, List2, List3) :-
findall(P, (nth0(P,List1,E), nth0(P,List2,E)), List3).
Differently from your solution, this one will work for any length lists...