Algorithm function for fibonacci series [closed]

Question

I'm not looking necessarily for an answer, but I am looking for what this question is asking of. Found this question studying for an interview but not sure what they're asking?

Write function that runs through the Fibonacci sequence and returns the index that is passed in as a parameter.

Answer 1

I interpret the question differently....given a number as an input, what is the index of that number in the series? e.g. input=5, then index is 5 (given the sequence is 0 1 1 2 3 5 where the index begins with 0)

This the code is as follows (which returns the index) [Disclaimer: Adapted from the code given at http://talkbinary.com/programming/c/fibonacci-in-c/ ]

int Fibonacci(int n)
{
  if ( n == 0 )
    return 0;
  if ( n== 1 )
    return 1;

  int fib1 = 0; 
  int fib2 = 1;
  int fib = 0;
  int i = 0;

for (i = 2; ; i++ ) 
{

    fib = fib1 + fib2;
    if ( n == fib )
       break;
    fib1 = fib2;
    fib2 = fib;
}


  return i;
}

Answer 2

public static int fibonacci(int i){
if(i==0)
  return 0;

if(i==1)
   return 1;
return fib(--i,0,1);
}


public static int fib(int num,int pre,int prepre){
   if(num==0){
    return prepre+pre;
   }
    return fib(--num,pre+prepre,pre);
}

Answer 3

What it seems to me is that you are asked to return the the nth fibonacci no., where n is the passed parameter. You can employ various methods to answer this question, whereas all these varies in time complexity and code complexity.

Method 1 ( Use recursion ) A simple method that is a direct recusrive implementation mathematical recurance relation given above.

int fib(int n)
{
    if ( n <= 1 )
    return n;
    return fib(n-1) + fib(n-2);
}

Time Complexity: T(n) = T(n-1) + T(n-2) which is exponential. We can observe that this implementation does a lot of repeated work (see the following recursion tree). So this is a bad implementation for nth Fibonacci number.

                     fib(5)   
                 /             \     
           fib(4)                fib(3)   
         /      \                /     \
     fib(3)      fib(2)         fib(2)    fib(1)
    /     \        /    \       /    \  

fib(2) fib(1) fib(1) fib(0) fib(1) fib(0) / \ fib(1) fib(0) Extra Space: O(n) if we consider the fuinction call stack size, otherwise O(1).

Method 2 ( Use Dynamic Programming ) We can avoid the repeated work done is the method 1 by storing the Fibonacci numbers calculated so far.

int fib(int n)
{
     /* Declare an array to store fibonacci numbers. */
      int f[n+1];
      int i;

     /* 0th and 1st number of the series are 0 and 1*/
     f[0] = 0;
     f[1] = 1;

    for (i = 2; i <= n; i++)
    {
       /* Add the previous 2 numbers in the series
        and store it */
       f[i] = f[i-1] + f[i-2];
    }

    return f[n];
}

Time Complexity: O(n) Extra Space: O(n)

Method 3 ( Space Otimized Method 2 ) We can optimize the space used in method 2 by storing the previous two numbers only because that is all we need to get the next Fibannaci number in series.

 int fib(int n)
 {
      int a = 0, b = 1, c, i;
      if( n == 0)
       return a;
      for (i = 2; i <= n; i++)
      {
        c = a + b;
        a = b;
       b = c;
    }
    return b;
  }

Time Complexity: O(n) Extra Space: O(1)

Method 4 ( Using power of the matrx {{1,1},{0,1}} ) This another O(n) which relies on the fact that if we n times multiply the matrix M = {{1,1},{0,1}} to itself (in other words calculate power(M, n )), then we get the (n+1)th Fibonacci number as the element at row and column (0, 0) in the resultant matrix.

The matrix representation gives the following closed expression for the Fibonacci numbers:

  /* Helper function that multiplies 2 matricies F and M of size 2*2, and
    puts the multiplication result back to F[][] */
  void multiply(int F[2][2], int M[2][2]);

  /* Helper function that calculates F[][] raise to the power n and puts the
    result in F[][]
    Note that this function is desinged only for fib() and won't work as general
    power function */
  void power(int F[2][2], int n);

  int fib(int n)
  {
    int F[2][2] = {{1,1},{1,0}};
    if(n == 0)
        return 0;
    power(F, n-1);

    return F[0][0];
  }

  void multiply(int F[2][2], int M[2][2])
  {
    int x =  F[0][0]*M[0][0] + F[0][1]*M[1][0];
    int y =  F[0][0]*M[0][1] + F[0][1]*M[1][1];
    int z =  F[1][0]*M[0][0] + F[1][1]*M[1][0];
    int w =  F[1][0]*M[0][1] + F[1][1]*M[1][1];

    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
  }

  void power(int F[2][2], int n)
  {
    int i;
    int M[2][2] = {{1,1},{1,0}};

    // n - 1 times multiply the matrix to {{1,0},{0,1}}
    for ( i = 2; i <= n; i++ )
        multiply(F, M);
  }

Time Complexity: O(n) Extra Space: O(1)

Method 5 ( Optimized Method 4 ) The method 4 can be optimized to work in O(Logn) time complexity. We can do recursive multiplication to get power(M, n) in the prevous method (Similar to the optimization done in this post)

  void multiply(int F[2][2], int M[2][2]);

  void power(int F[2][2], int n);

  /* function that returns nth Fibonacci number */
  int fib(int n)
  {
    int F[2][2] = {{1,1},{1,0}};
    if(n == 0)
      return 0;
    power(F, n-1);
    return F[0][0];
  }

  /* Optimized version of power() in method 4 */
  void power(int F[2][2], int n)
  {
    if( n == 0 || n == 1)
        return;
    int M[2][2] = {{1,1},{1,0}};

    power(F, n/2);
    multiply(F, F);

    if( n%2 != 0 )
       multiply(F, M);
  }

  void multiply(int F[2][2], int M[2][2])
  {
    int x =  F[0][0]*M[0][0] + F[0][1]*M[1][0];
    int y =  F[0][0]*M[0][1] + F[0][1]*M[1][1];
    int z =  F[1][0]*M[0][0] + F[1][1]*M[1][0];
    int w =  F[1][0]*M[0][1] + F[1][1]*M[1][1];

    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
  }

Time Complexity: O(Logn) Extra Space: O(Logn) if we consider the function call stack size, otherwise O(1).

Driver Program: int main() { int n = 9; printf("%d", fib(9)); getchar(); return 0; }

References: http://en.wikipedia.org/wiki/Fibonacci_number http://www.ics.uci.edu/~eppstein/161/960109.html

Answer 4

firstly,you can update your base math information about Fibonacci with this link from wiki. and look at this formula for fast calculate.and you can read all of about it in this link.

This is recursive function to compute nth Fibonacci number and is of O(2^n) time:

 int Fibonacci(int n) {  
        if (n == 0 || n == 1)  return n;
        else
        return Fibonacci(n - 1) + Fibonacci(n - 2); }

Computing the Sequence

You might argue that in terms of actually computing the values of the Fibonacci sequence on a computer, you’re better off using the original recurrence relation, f[n]=f[n−1]+f[n−2]. I’m inclined to agree. To use the direct closed-form solution for large n, you need to maintain a lot of precision. Even with 9 decimal places out, fn≈round(0.723606798⋅(1.618033989)n), for example, is only valid for up to n=38 (observe here versus here). Also, adding integers is much less computationally expensive and more precise than exponentiating a symbolic fraction or a floating point value

this is better idea to compute nth Fibonacci number and is of O(n) time:

int Fibonacci(int n) { 
if(n <= 0) return 0;
if(n > 0 && n < 3) return 1;

int result = 0;
int preOldResult = 1;
int oldResult = 1;

for (int i=2;i<n;i++) { 
    result = preOldResult + oldResult;
    preOldResult = oldResult;
    oldResult = result;
}

return result;}

and this is the best way to compute nth Fibonacci number and is of O(log(n)) time:

this link:

As you are already suspecting, this will work very similar. Use the n-th power of the x * x matrix

|1 0 0 0  .... 1 1|
|1 
|  1
|    1
|      1
|        1
...................
...................
|          ... 1 0|

This is easy to understand if you multiply this matrix with the vector

f(n-1), f(n-2), ... , f(n-x+1), f(n-x)

which results in

f(n), f(n-1), ... , f(n-x+1)

Matrix exponentiation can be done in O(log(n)) time (when x is considered to be constant).

For the Fibonacci recurrence, there is also a closed formula solution, see here http://en.wikipedia.org/wiki/Fibonacci_number, look for Binet's or Moivre's formula.

and look at: 1-nth fibonacci number in sublinear time

Answer 5

It's a very poorly worded question, but you have to assume they are asking for the n^th Fibonnaci number where n is provided as the parameter.

In addition to all the techniques listed by others, for n > 1 you can also use the golden ratio method, which is quicker than any iterative method. But as the question says 'run through the Fibonacci sequence' this may not qualify. You'd probably also scare them to death.