fizzBuzz.fs

let fizzBuzz attempt =
    match attempt with
    | (0, 0, _) -> "FizzBuzz"
    | (0, _, _) -> "Fizz"
    | (_, 0, _) -> "Buzz"
    | (_, _, n) -> n.ToString()

for num in 1..100 do
    printfn "%s" (fizzBuzz (num % 3, num % 5, num))
    
(*

    fizzBuzz takes a single attempt.
    
    An attempt is a 3-element tuple from a number from 1 to 100.
    
    The 3-tuple structure looks like this:
    
    1. The remainder of the number and 3 (to represent if it's divisible by 3)
    2. The remainder of the number and 5 (to represent if it's dibisible by 5)
    3. The number itself
    
    So, with the first number in 1 through 100, we have 1,
    which would have an attempt that looks like this: (1, 1, 1)
    
    The remainder of 1 and 3 is 1, the remainder of 1 and 5 is 1, and the number itself is 1
    
    Skipping ahead to 3 would look like this: (0, 3, 3)
    The remainder of 3 and 3 is 0, the remainder of 3 and 5 is 3, and the number itself is 3
    
    For 5: (2, 0, 5)
    The remainder of 5 and 3 is 2, the remainder of 5 and 5 is 0, and the number itself is 5
    
    Skipping ahead to 15: (0, 0, 15)
    The remainder of 15 and 3 is 0, the remainder of 15 and 5 is 0, and the number itself is 15
    
    The fizzBuzz function takes this 3-tuple and matches it against 4 cases:
    (0, 0, _) -> "FizzBuzz"
    (0, _, _) -> "Fizz"
    (_, 0, _) -> "Buzz"
    (_, _, n) -> n
    
    The case of (0, 0, _) says "this number is divisible by both 3 and 5" and therefore we show "FizzBuzz"
    (0, _, _) says "this number is not divisible by both 3 and 5, but is divisble by 3", so we show "Fizz"
    (_, 0, _) says the same, but for its divisibility by 5
    (_, _, n) says that it's not divisible by 3 or 5, so we should just show the number itself

*)