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Asked: 2026-05-22T03:01:08+00:00

How would the Fibonacci’s closed form code look like in haskell ?

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How would the Fibonacci’s closed form code look like in haskell?

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  1. Editorial Team

    Trivially, Binet’s formula, from the Haskell wiki page is given in Haskell as:

    fib n = round $ phi ^ n / sq5
  where
    sq5 = sqrt 5 
    phi = (1 + sq5) / 2

    Which includes sharing of the result of the square root. For example:

    *Main> fib 1000
4346655768693891486263750038675
5014010958388901725051132915256
4761122929200525397202952340604
5745805780073202508613097599871
6977051839168242483814062805283
3118210513272735180508820756626
59534523370463746326528

    For arbitrary integers, you’ll need to be a bit more careful about the conversion to floating point values.
    Note that Binet’s value differs from the recursive formula by quite a bit at this point:

    *Main> let fibs = 0 : 1 : zipWith (+) fibs (tail fibs) 
*Main> fibs !!   1000
4346655768693745643568852767504
0625802564660517371780402481729
0895365554179490518904038798400
7925516929592259308032263477520
9689623239873322471161642996440
9065331879382989696499285160037
04476137795166849228875

    You may need more precision 🙂

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