An iterative algorithm for Fibonacci numbers

Question

I am interested in an iterative algorithm for Fibonacci numbers, so I found the formula on wiki...it looks straight forward so I tried it in Python...it doesn\'t have a prob

Answer 1

Assuming these values for the fibonacci sequence:

F(0) = 0;

F(1) = 1;

F(2) = 1;

F(3) = 2

For values of N > 2 we'll calculate the fibonacci value with this formula:

F(N) = F(N-1) + F(N-2)

One iterative approach we can take on this is calculating fibonacci from N = 0 to N = Target_N, as we do so we can keep track of the previous results of fibonacci for N-1 and N-2

public int Fibonacci(int N)
{
    // If N is zero return zero
    if(N == 0)
    {
        return 0;
    }

    // If the value of N is one or two return 1
    if( N == 1 || N == 2)
    {
       return 1;
    }

    // Keep track of the fibonacci values for N-1 and N-2
    int N_1 = 1;
    int N_2 = 1;

    // From the bottom-up calculate all the fibonacci values until you 
    // reach the N-1 and N-2 values of the target Fibonacci(N)
    for(int i =3; i < N; i++)
    {
       int temp = N_2;
       N_2 = N_2 + N_1;
       N_1 = temp;
    }

    return N_1 + N_2; 
}

Answer 2

I came across this on another thread and it is significantly faster than anything else I have tried and wont time out on large numbers. Here is a link to the math.

def fib(n):
    v1, v2, v3 = 1, 1, 0  
    for rec in bin(n)[3:]: 
        calc = v2*v2
        v1, v2, v3 = v1*v1+calc, (v1+v3)*v2, calc+v3*v3
        if rec=='1':    v1, v2, v3 = v1+v2, v1, v2
    return v2

Answer 3

This work (intuitively)

def fib(n):
    if n < 2:
        return n
    o,i = 0,1
    while n > 1:
        g = i
        i = o + i
        o = g
        n -= 1
    return i

Answer 4

How about this simple but fastest way... (I just discovered!)

def fib(n):
    x = [0,1]
    for i in range(n >> 1):
        x[0] += x[1]
        x[1] += x[0]
    return x[n % 2]

Note! as a result, this simple algorithm only uses 1 assignment and 1 addition, since loop length is shorten as 1/2 and each loop includes 2 assignment and 2 additions.

Answer 5

You are returning a value within a loop, so the function is exiting before the value of y ever gets to be any more than 1.

If I may suggest something shorter, and much more pythonful:

def fibs(n):                                                                                                 
    fibs = [0, 1, 1]                                                                                           
    for f in range(2, n):                                                                                      
        fibs.append(fibs[-1] + fibs[-2])                                                                         
    return fibs[n]

This will do exactly the same thing as your algorithm, but instead of creating three temporary variables, it just adds them into a list, and returns the nth fibonacci number by index.

Answer 6

Non recursive Fibonacci sequence in python

def fibs(n):
    f = []
    a = 0
    b = 1
    if n == 0 or n == 1:
        print n
    else:
        f.append(a)
        f.append(b)
        while len(f) != n:
            temp = a + b
            f.append(temp)
            a = b
            b = temp

    print f

fibs(10)

Output: [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]