Fibonacci in Prolog - Breaks if False - prolog

Question

I have a program fib(X,Y). If Y is the Xth Fibonacci number it returns True else it should return False. My program breaks anytime I input statement which is false.
fib(R,V) :- fib(0,1,R,V).
fib(X, Y, 0, V) :- Y == V.
fib(X, Y, R, V) :- Z is X + Y, C is R - 1, fib(Y, Z, C, V).
fib(0,1) -> True
fib(1,1) -> True
fib(2,2) -> True
fib(3,3) -> True
fib(4,5) -> True
fib(3,5) -> Won't finish.
What do I do wrong? I am using https://swish.swi-prolog.org/ to run my program queries.

Answer 1

The problem here is that you write two clauses fib(X, Y, 0, V) :- and fib(X, Y, R, V) :-. Prolog uses backtracking: in case one clause has been tried, it wil - regardless of sucess or failure - also later retry the next clause (there are some meta-predicates like once/1 that can alter this).
So even if R is 0, or lower, Prolog will also try the second clause.
A quick way to fix this is by using a guards for the second clause:
fib(_, Y, 0, V) :-
Y == V.
fib(X, Y, R, V) :-
R > 0,
Z is X + Y,
C is R - 1,
fib(Y, Z, C, V).
Furthermore your code is not very elegant in the sense that you can not use the relation in a reversed way, nor can we query for the X-th element.
For instance you use Y == V, but this blocks unification: if we want to know the X-th fibonacci number, we want a way to propagate the result back. So we can use unification instead:
fib(_, V, 0, V).
fib(X, Y, R, V) :-
R > 0,
Z is X + Y,
C is R - 1,
fib(Y, Z, C, V).
But now we still do not have a bidirectional relation: we can not obtain the X for a given value V. This is more complex. The easiest way is probably using clpfd for this:
:- use_module(library(clpfd)).
fib(_, V, 0, V).
fib(X, Y, R, V) :-
R #> 0,
V #>= Y,
Z is X + Y,
C #= R - 1,
fib(Y, Z, C, V).
Now we can:
enumerate all indices and the corresponding Fibonacci numbers:
?- fib(A,B).
A = 0,
B = 1 ;
A = B, B = 2 ;
A = B, B = 3 ;
A = 4,
B = 5 ;
A = 5,
B = 8 ;
A = 6,
B = 13 ;
A = 7,
B = 21
...
Obtain the i-th Fibonacci number:
?- fib(2,B).
B = 2 ;
false.
?- fib(10,B).
B = 89 ;
false.
obtain the i for which the corresponding Fibonacci number is a certain value:
?- fib(A,1).
A = 0 ;
A = 1 ;
false.
?- fib(A,2).
A = 2 ;
false.
?- fib(A,3).
A = 3 ;
false.
?- fib(A,4).
false.
?- fib(A,5).
A = 4 ;
false.
Check if the i-th Fibonacci number is a given value:
?- fib(4,5).
true ;
false.
?- fib(4,6).
false.
?- fib(4,10).
false.
?- fib(5,8).
true ;
false.