Making Fibonacci faster [duplicate]

Answer 1

• Fibonacci(0) = 0
• Fibonacci(1) = 1
• Fibonacci(n) = Fibonacci(n - 1) + Fibonacci(n - 2), when n >= 2

Usually there are 2 ways to calculate Fibonacci number:

1. Recursion:

   public long getFibonacci(long n) {
     if(n <= 1) {
       return n;
     } else {
       return getFibonacci(n - 1) + getFibonacci(n - 2);
     }
   }
   

   This way is intuitive and easy to understand, while because it does not reuse calculated Fibonacci number, the time complexity is about O(2^n), but it does not store calculated result, so it saves space a lot, actually the space complexity is O(1).

2. Dynamic Programming:

   public long getFibonacci(long n) {
     long[] f = new long[(int)(n + 1)];
     f[0] = 0;
     f[1] = 1;
     for(int i=2;i<=n;i++) {
       f[i] = f[i - 1] + f[i - 2];
     }
     return f[(int)n];
   }
   

   This Memoization way calculated Fibonacci numbers and reuse them when calculate next one. The time complexity is pretty good, which is O(n), while space complexity is O(n). Let's investigate whether the space complexity can be optimized... Since f(i) only requires f(i - 1) and f(i - 2), there is not necessary to store all calculated Fibonacci numbers.

   The more efficient implementation is:

   public long getFibonacci(long n) {
     if(n <= 1) {
       return n;
     }
     long x = 0, y = 1;
     long ans;
     for(int i=2;i<=n;i++) {
       ans = x + y;
       x = y;
       y = ans;
     }
     return ans;
   }
   

   With time complexity O(n), and space complexity O(1).

Added: Since Fibonacci number increase amazing fast, long can only handle less than 100 Fibonacci numbers. In Java, we can use BigInteger to store more Fibonacci numbers.

Answer 2

Here's a way of provably doing it in O(log n) (as the loop runs log n times):

/* 
 * Fast doubling method
 * F(2n) = F(n) * (2*F(n+1) - F(n)).
 * F(2n+1) = F(n+1)^2 + F(n)^2.
 * Adapted from:
 *    https://www.nayuki.io/page/fast-fibonacci-algorithms
 */
private static long getFibonacci(int n) {
    long a = 0;
    long b = 1;
    for (int i = 31 - Integer.numberOfLeadingZeros(n); i >= 0; i--) {
        long d = a * ((b<<1) - a);
        long e = (a*a) + (b*b);
        a = d;
        b = e;
        if (((n >>> i) & 1) != 0) {
            long c = a+b;
            a = b;
            b = c;
        }
    }
    return a;
}

I am assuming here (as is conventional) that one multiply / add / whatever operation is constant time irrespective of number of bits, i.e. that a fixed-length data type will be used.

This page explains several methods of which this is the fastest. I simply translated it away from using BigInteger for readability. Here's the BigInteger version:

/* 
 * Fast doubling method.
 * F(2n) = F(n) * (2*F(n+1) - F(n)).
 * F(2n+1) = F(n+1)^2 + F(n)^2.
 * Adapted from:
 *    http://www.nayuki.io/page/fast-fibonacci-algorithms
 */
private static BigInteger getFibonacci(int n) {
    BigInteger a = BigInteger.ZERO;
    BigInteger b = BigInteger.ONE;
    for (int i = 31 - Integer.numberOfLeadingZeros(n); i >= 0; i--) {
        BigInteger d = a.multiply(b.shiftLeft(1).subtract(a));
        BigInteger e = a.multiply(a).add(b.multiply(b));
        a = d;
        b = e;
        if (((n >>> i) & 1) != 0) {
            BigInteger c = a.add(b);
            a = b;
            b = c;
        }
    }
    return a;
}