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Fibonacci Without Loops or Recursion

, 17 Dec 2012 CPOL
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A method for calculating a Fibonacci number without using loops or recursion.


While reading one of our Insider News posts which linked to Evan Miller's site,  he mentioned a mathematical means of producing a Fibonacci number without using loops or recursion.   I decided to post the C# version of it here, but in no way do I claim credit to creating this.   I thought it was interesting enough to share for those who might not read the Insider News articles.  

You can read more about this closed-form solution on wiki.

The Code

public static long Fibonacci(long n)
    return (long)Math.Round(0.44721359549995682d * Math.Pow(1.6180339887498949d, n));

NOTE: Due to limits of precision, the preceding formula is only accurate up to n = 77. 


Based on YvesDaoust's recommendation, I've updated the formula to use a simpler version of the closed form solution (also found on Wiki), as it proves to be faster and more compact.

Furthermore, I've adjusted the constants slightly to improve the function's accuracy.


This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)


About the Author

Andrew Rissing
Software Developer (Senior)
United States United States
Since I've begun my profession as a software developer, I've learned one important fact - change is inevitable. Requirements change, code changes, and life changes.
So..If you're not moving forward, you're moving backwards.

Comments and Discussions

GeneralThoughts PinprofessionalPIEBALDconsult8-Sep-13 14:55 
GeneralRe: Thoughts PinmemberAndrew Rissing10-Sep-13 6:08 
Questionformula's missing part PinmemberMember 33856988-Sep-13 12:30 
AnswerRe: formula's missing part PinmemberAndrew Rissing10-Sep-13 6:06 
GeneralRe: formula's missing part PinmemberYuksel YILDIRIM14-Oct-13 16:56 
GeneralMy vote of 2 PinmemberYvesDaoust17-Dec-12 2:57 
QuestionCan do better PinmemberYvesDaoust17-Dec-12 2:51 
AnswerRe: Can do better PinmemberAndrew Rissing17-Dec-12 5:48 
AnswerRe: Can do better PinmemberAndrew Rissing17-Dec-12 6:04 
GeneralRe: Can do better [modified] PinmemberYvesDaoust17-Dec-12 6:32 
GeneralRe: Can do better PinmemberAndrew Rissing17-Dec-12 13:04 
GeneralRe: Can do better PinmemberYvesDaoust17-Dec-12 21:34 
GeneralRe: Can do better PinmemberAndrew Rissing18-Dec-12 5:24 
GeneralMy vote of 5 PinmemberManish Choudhary .NET expert14-Dec-12 19:50 
BugThis is correct only for n up to 70 PinmemberMatt T Heffron13-Dec-12 13:12 
This is correct only for the first 70 Fibonacci numbers.
At number 71 the closed form is 1 too big:
308061521170130 closed form
308061521170129 loop
Presumably, this is due to rounding errors and precision becoming significant.
Changing the Math.Round() to Math.Floor() only gets you one additional correct value.
Also, on average over computing the 70 values, the closed form is about 25% slower.
  using System;
  using System.Diagnostics;
  class Program
    static void Main(string[] args)
      Stopwatch swCF = new Stopwatch();
      Stopwatch swLoop = new Stopwatch();
      const int repeats = 10000;
      for (int rep = 0; rep < repeats; rep++)
        for (long i = 1; i <= 100; i++)
          long cf = FibonacciCF(i);
          long lp = FibonacciLoop(i);
          if (cf != lp)
            if (rep == 0)
              // only report this ONCE!!
              Console.WriteLine("mismatch at {0}: cf={1} lp={2}", i, cf, lp);
      Console.WriteLine("Repeats = {0}", repeats);
      Console.WriteLine("Closed form time: {0}", TimeSpan.FromTicks(swCF.ElapsedTicks));
      Console.WriteLine("Loop time: {0}", TimeSpan.FromTicks(swLoop.ElapsedTicks));
      Console.WriteLine("Press Enter to complete...");
    // compute sqrt(5) only once, outside of timing.
    private static readonly double SquareRootOf5 = Math.Sqrt(5.0d);
    public static long FibonacciCF(long n)
      //double SquareRootOf5 = Math.Sqrt(5.0d);
      return (long)Math.Round((
          (Math.Pow(0.5d + 0.5d * SquareRootOf5, n) -
          Math.Pow(0.5d - 0.5d * SquareRootOf5, n)) /
    public static long FibonacciLoop(long n)
      long prev = 0;
      long curr = 1;
      for (long i = 1; i < n; i++)
        long oldprev = prev;
        prev = curr;
        curr += oldprev;
      return curr;

GeneralRe: This is correct only for n up to 70 PinmemberAndrew Rissing13-Dec-12 13:45 
GeneralRe: This is correct only for n up to 70 PinmemberAndrew Rissing17-Dec-12 13:05 

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