General FizzBuzz: Difference between revisions

=={{header|Prolog}}==

Assuming the user specifies as input two facts of the form:

<lang prolog>maxNumber(105).

factors([(3, "Fizz"), (5, "Buzz"), (7, "Baxx")]).</lang>

A simple Prolog solution to the generalised FizzBuzz problem is as follows:

<lang prolog>fizzbuzz(N,Res) :-

factors(Fs), fizzbuzz(N,Fs,S),

( (S = "", Res is N) ; Res = S ).

fizzbuzz(_,[],"").

fizzbuzz(N,[(F,S)|Fs],Res) :-

fizzbuzz(N,Fs,OldRes),

( N mod F =:= 0, string_concat(S,OldRes,Res) ; Res = OldRes ).

loop(B,B).

loop(A,B) :- A < B, fizzbuzz(A,Res), writeln(Res), Next is A+1, loop(Next,B).

go :- maxNumber(M), MLast is M+1, loop(1,MLast).</lang>

It is worth noting that

<lang prolog>factors([(3, "Fizz"), (5, "Buzz")]).</lang>

corresponds to basic FizzBuzz and that the proposed solution can handle an arbitrary number of factors.