Python solution that uses the fact that the ratio of consecutive numbers in the sequence approaches Phi
def fib(n): if n < 3: return 1 return int(round(1.618 * fib(n-1)))
This one is absolutely beautiful.
I tried running the code but when I tried to run fib(10) it returns 1, did i do something wrong?
1
Nope, I made a stupid mistake when writing this function on Dev.to, I originally wrote it in the REPL. Fixing it now.
Excellent stuff!
This is closely related to my closed form solution, as they are both based on Binet's Formula
Using a hard-coded value, mine can be reduced as well:
import math sq5 = math.sqrt(5.0) φ = (1.0 + sq5)/2.0 ψ = -1/φ def fib_x( n ): return int( (φ ** n - ψ ** n) / sq5 ) for i in range(20): print(fib_x(i))
Where φ is the Phi value you are using.
φ
Wow, so we can actually get O(1) with this? Amazing
It s not actually O(1) since ** n takes time too :))
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Python solution that uses the fact that the ratio of consecutive numbers in the sequence approaches Phi
This one is absolutely beautiful.
I tried running the code but when I tried to run fib(10) it returns
1, did i do something wrong?Nope, I made a stupid mistake when writing this function on Dev.to, I originally wrote it in the REPL. Fixing it now.
Excellent stuff!
This is closely related to my closed form solution, as they are both based on Binet's Formula
Using a hard-coded value, mine can be reduced as well:
Where
φis the Phi value you are using.Wow, so we can actually get O(1) with this? Amazing
It s not actually O(1) since ** n takes time too :))