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  1. Prepare
  2. Functional Programming
  3. Recursion
  4. Fibonacci Numbers
  5. Discussions

Fibonacci Numbers

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  • dgmatos30
     + 0 comments

    That's a simple haskell solution which matches the fibonacci formula:

    fib(n) Sequence
    Nth: 1 2 3 4 5 6 7 08 09 10 11 12 [...]
    Terms: 0 1 1 2 3 5 8 13 21 34 55 89 [...] 

    fib 1 = 0
fib 2 = 1
fib n = fib(n-1) + fib(n-2)

  • AlsMrqs
     + 0 comments
    fib n = f 0 1 n
    where
    f a b n
        | n == 1    = a
        | otherwise = f b (a+b) $ n-1

  • trisye
     + 0 comments

    Why use this (wrong) choice of indices? The Fibonacci numbers must start with f(0) = 0, f(1) = 1, if only for aesthetic reasons, like the gcd-property. Thanks for the video link btw.

  • jesse_schultz2
     + 0 comments

    for F#

    here is my solution:

    let rec getFib(n:int) = 
    if n < 4
        then 1 
    else 
        getFib (n - 1) + getFib (n - 2)
    
let main =
    let n = System.Console.ReadLine() |> int
    let ret = getFib(n)
    printfn "%d" ret

  • scriptori
     + 0 comments

    Kotlin

        fun fib(n: Int): Int {
        return when (n) {
            0 -> 0
            1 -> 1
            else -> fib(n - 1) + fib(n - 2)
        }
    }
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