Optimization of Fibonacci sequence generating algorithm

Question

As we all know, the simplest algorithm to generate Fibonacci sequence is as follows:

if(n<=0) return 0;
else if(n==1) return 1;
f(n) = f(n-1) + f(n-2);

But this algorithm has some repetitive calculation. For example, if you calculate f(5), it will calculate f(4) and f(3). When you calculate f(4), it will again calculate both f(3) and f(2). Could someone give me a more time-efficient recursive algorithm?

Answer 1

I have read about some of the methods for calculating Fibonacci with efficient time complexity following are some of them -

Method 1 - Dynamic Programming Now here the substructure is commonly known hence I'll straightly Jump to the solution -

static int fib(int n) 
{ 
int f[] = new int[n+2]; // 1 extra to handle case, n = 0 
int i; 

f[0] = 0; 
f[1] = 1; 

for (i = 2; i <= n; i++) 
{ 
    f[i] = f[i-1] + f[i-2]; 
} 

return f[n]; 
}

A space-optimized version of above can be done as follows -

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

Method 2- ( Using power of the matrix {{1,1},{1,0}} )

This an O(n) which relies on the fact that if we n times multiply the matrix M = {{1,1},{1,0}} 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. This solution would have O(n) time.

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

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

return F[0][0]; 
} 

/*multiplies 2 matrices F and M of size 2*2, and 
puts the multiplication result back to F[][] */
static void multiply(int F[][], int M[][]) 
{ 
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; 
} 

/*function that calculates F[][] raise to the power n and puts the 
result in F[][]*/
static void power(int F[][], int n) 
{ 
int i; 
int M[][] = new int[][]{{1,1},{1,0}}; 

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

This can be optimized to work in O(Logn) time complexity. We can do recursive multiplication to get power(M, n) in the previous method.

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

return F[0][0]; 
} 

static void multiply(int F[][], int M[][]) 
{ 
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; 
} 

static void power(int F[][], int n) 
{ 
if( n == 0 || n == 1) 
  return; 
int M[][] = new int[][]{{1,1},{1,0}}; 

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

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

Method 3 (O(log n) Time) Below is one more interesting recurrence formula that can be used to find nth Fibonacci Number in O(log n) time.

If n is even then k = n/2: F(n) = [2*F(k-1) + F(k)]*F(k)

If n is odd then k = (n + 1)/2 F(n) = F(k)*F(k) + F(k-1)*F(k-1) How does this formula work? The formula can be derived from the above matrix equation. fibonaccimatrix

Taking determinant on both sides, we get (-1)n = Fn+1Fn-1 – Fn2 Moreover, since AnAm = An+m for any square matrix A, the following identities can be derived (they are obtained from two different coefficients of the matrix product)

FmFn + Fm-1Fn-1 = Fm+n-1

By putting n = n+1,

FmFn+1 + Fm-1Fn = Fm+n

Putting m = n

F2n-1 = Fn2 + Fn-12

F2n = (Fn-1 + Fn+1)Fn = (2Fn-1 + Fn)Fn (Source: Wiki)

To get the formula to be proved, we simply need to do the following If n is even, we can put k = n/2 If n is odd, we can put k = (n+1)/2

public static int fib(int n) 
{ 

    if (n == 0) 
        return 0; 

    if (n == 1 || n == 2) 
        return (f[n] = 1); 

    // If fib(n) is already computed 
    if (f[n] != 0) 
        return f[n]; 

    int k = (n & 1) == 1? (n + 1) / 2 
                        : n / 2; 

    // Applyting above formula [See value 
    // n&1 is 1 if n is odd, else 0. 
    f[n] = (n & 1) == 1? (fib(k) * fib(k) +  
                    fib(k - 1) * fib(k - 1)) 
                   : (2 * fib(k - 1) + fib(k))  
                   * fib(k); 

    return f[n]; 
} 

Method 4 - Using a formula In this method, we directly implement the formula for the nth term in the Fibonacci series. Time O(1) Space O(1) Fn = {[(√5 + 1)/2] ^ n} / √5

static int fib(int n) { 
double phi = (1 + Math.sqrt(5)) / 2; 
return (int) Math.round(Math.pow(phi, n)  
                    / Math.sqrt(5)); 
} 

Reference: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html

Answer 2

Look here for implementation in Erlang which uses formula . It shows nice linear resulting behavior because in O(M(n) log n) part M(n) is exponential for big numbers. It calculates fib of one million in 2s where result has 208988 digits. The trick is that you can compute exponentiation in O(log n) multiplications using (tail) recursive formula (tail means with O(1) space when used proper compiler or rewrite to cycle):

% compute X^N
power(X, N) when is_integer(N), N >= 0 ->
    power(N, X, 1).

power(0, _, Acc) ->
    Acc;
power(N, X, Acc) ->
    if N rem 2 =:= 1 ->
            power(N - 1,   X,     Acc * X);
        true ->
            power(N div 2, X * X, Acc)
    end.

where X and Acc you substitute with matrices. X will be initiated with and Acc with identity I equals to .