C# fibonacci function returning errors

Question

I am practising a C# console application, and I am trying to get the function to verify if the number appears in a fibonacci series or not but I'm getting errors.

What I did was:

class Program
{
    static void Main(string[] args)
    {
        System.Console.WriteLine(isFibonacci(20));
    }
    static int isFibonacci(int n)
    {
        int[] fib = new int[100];
        fib[0] = 1;
        fib[1] = 1;
        for (int i = 2; i <= 100; i++)
        {
            fib[i] = fib[i - 1] + fib[i - 2];

            if (n == fib[i])
            {
                return 1;
            }



        }
        return 0;
    }
}

Can anybody tell me what am I doing wrong here?

Answer 1

And here is a solution that beats all of yours!

Because, why iteration when you have smart mathematicians doing closed-form solutions for you? :)

static bool IsFibonacci(int number)
{
    //Uses a closed form solution for the fibonacci number calculation.
    //http://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression

    double fi = (1 + Math.Sqrt(5)) / 2.0; //Golden ratio
    int n = (int) Math.Floor(Math.Log(number * Math.Sqrt(5) + 0.5, fi)); //Find's the index (n) of the given number in the fibonacci sequence

    int actualFibonacciNumber = (int)Math.Floor(Math.Pow(fi, n) / Math.Sqrt(5) + 0.5); //Finds the actual number corresponding to given index (n)

    return actualFibonacciNumber == number;
}

Answer 2

Here's a fun solution using an infinite iterator block:

IEnumerable<int> Fibonacci()
{
   int n1 = 0;
   int n2 = 1;

   yield return 1;
   while (true)
   {
      int n = n1 + n2;
      n1 = n2;
      n2 = n;
      yield return n;
   }
}

bool isFibonacci(int n)
{
    foreach (int f in Fibonacci())
    {
       if (f > n) return false;
       if (f == n) return true;
    }
}

I actually really like this kind of Fibonacci implementation vs the tradition recursive solution, because it keeps the work used to complete a term available to complete the next. The traditional recursive solution duplicates some work, because it needs two recursive calls each term.

Answer 3

The problem lies in <= the following statement:

for (int i = 2; i <= 100; i++)

more to the point the =. There is no fib[100] (C# zero counts) so when you check on i=100 you get an exception.

the proper statement should be

for (int i = 2; i < 100; i++)

or even better

for (int i = 2; i < fib.Length; i++)

Answer 4

Well, for starters your array is only 10 long and you're filling it with ~100 items (out-of-range-exception) - but there are better ways to do this...

for example, using this post:

long val = ...
bool isFib = Fibonacci().TakeWhile(x => x <= val).Last() == val;

Answer 5

int[] fib = new int[10];
for (int i = 2; i <= *100*; i++)

You're going out of the bounds of your array because your loop conditional is too large. A more traditional approach would be to bound the loop by the size of the array:

for (int i = 2; i < fib.Length; i++)

And make your array bigger, but as Marc said, there are better ways to do this, and I would advise you spend some time reading the wikipedia article on Fibonacci numbers.

Answer 6

One thing you can do is check for an early exit. Since you're trying to determine if a given number is in the Fibonacci sequence, you can do bounds checking to exit early.

Example:

static bool isFibonacci(int n)
{
    int[] fib = new int[100];
    fib[0] = 1;
    fib[1] = 1;
    for (int i = 2; i <= fib.Length; i++)
    {
        fib[i] = fib[i - 1] + fib[i - 2];

        if (n == fib[i])
        {
            return true;
        }
        else if (n < fib[i])
        {
            return false;  //your number has been surpassed in the fib seq
        }
    }
    return false;
}

Answer 7

public static int FibNo(int n) {
    int result = 0; int No = 0; int N1 = 1;

    if (n< 0)
    { throw new ArguementException("number must be a positive value"); }

    if (n <= 1) 
    { result = n; return result; }

    for(int x=1; x < n; x++) 
    { result = No + N1; No = N1; N1=result; }

    return result;

}