Making Fibonacci faster [duplicate]

Answer 1

If you need to compute n th fibonacci numbers very frequently I suggest using amalsom's answer.

But if you want to compute a very big fibonacci number, you will run out of memory because you are storing all smaller fibonacci numbers. The following pseudocode only keeps the last two fibonacci numbers in memory, i.e. it requires much less memory:

fibonacci(n) {
    if n = 0: return 0;
    if n = 1: return 1;
    a = 0;
    b = 1;
    for i from 2 to n: {
        sum = a + b;
        a = b;
        b = sum;
    }
    return b;
}

Analysis
This can compute very high fibonacci numbers with quite low memory consumption: We have O(n) time as the loop repeats n-1 times. The space complexity is interesting as well: The nth fibonacci number has a length of O(n), which can easily be shown:
F_n <= 2 * F_n-1
Which means that the nth fibonacci number is at most twice as big as its predecessor. Doubling a number in binary is equivalent with a single left-shift, which increases the number of necessary bits by one. So representing the nth fibonacci number takes at most O(n) space. We have at most three successive fibonacci numbers in memory which makes O(n) + O(n-1) + O(n-2) = O(n) total space consumption. In contrast to this the memoization algorithm always keeps the first n fibonacci numbers in memory, which makes O(n) + O(n-1) + O(n-2) + ... + O(1) = O(n^2) space consumption.

So which way should one use?
The only reason to keep all lower fibonacci numbers in memory is if you need fibonacci numbers very frequently. It is a question of balancing time with memory consumption.

Answer 2

Here is snipped of code with iterative approach instead of recursion.

Output example:

Enter n: 5
F(5) = 5 ... computed in 1 milliseconds
Enter n: 50
F(50) = 12586269025 ... computed in 0 milliseconds
Enter n: 500
F(500) = ...4125 ... computed in 2 milliseconds
Enter n: 500
F(500) = ...4125 ... computed in 0 milliseconds
Enter n: 500000
F(500000) = ...453125 ... computed in 5,718 milliseconds
Enter n: 500000
F(500000) = ...453125 ... computed in 0 milliseconds

Some pieces of results are omitted with ... for better view.

Code snippet:

public class CachedFibonacci {
    private static Map<BigDecimal, BigDecimal> previousValuesHolder;
    static {
        previousValuesHolder = new HashMap<>();
        previousValuesHolder.put(BigDecimal.ZERO, BigDecimal.ZERO);
        previousValuesHolder.put(BigDecimal.ONE, BigDecimal.ONE);
    }

    public static BigDecimal getFibonacciOf(long number) {
        if (0 == number) {
            return BigDecimal.ZERO;
        } else if (1 == number) {
            return BigDecimal.ONE;
        } else {
            if (previousValuesHolder.containsKey(BigDecimal.valueOf(number))) {
                return previousValuesHolder.get(BigDecimal.valueOf(number));
            } else {
                BigDecimal olderValue = BigDecimal.ONE,
                        oldValue = BigDecimal.ONE,
                        newValue = BigDecimal.ONE;

                for (int i = 3; i <= number; i++) {
                    newValue = oldValue.add(olderValue);
                    olderValue = oldValue;
                    oldValue = newValue;
                }
                previousValuesHolder.put(BigDecimal.valueOf(number), newValue);
                return newValue;
            }
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (true) {
            System.out.print("Enter n: ");
            long inputNumber = scanner.nextLong();
            if (inputNumber >= 0) {
                long beginTime = System.currentTimeMillis();
                BigDecimal fibo = getFibonacciOf(inputNumber);
                long endTime = System.currentTimeMillis();
                long delta = endTime - beginTime;

                System.out.printf("F(%d) = %.0f ... computed in %,d milliseconds\n", inputNumber, fibo, delta);
            } else {
                System.err.println("You must enter number > 0");
                System.out.println("try, enter number again, please:");
                break;
            }
        }
    }
}

This approach runs much faster than the recursive version.

In such a situation, the iterative solution tends to be a bit faster, because each recursive method call takes a certain amount of processor time. In principle, it is possible for a smart compiler to avoid recursive method calls if they follow simple patterns, but most compilers don’t do that. From that point of view, an iterative solution is preferable.

Answer 3

Get away from the Fibonacci recursion and use the identities

(F(2n), F(2n-1)) = (F(n)^2 + 2 F(n) F(n-1), F(n)^2+F(n-1)^2)
(F(2n+1), F(2n)) = (F(n+1)^2+F(n)^2, 2 F(n+1) F(n) - F(n)^2)

This allows you to compute (F(m+1), F(m)) in terms of (F(k+1), F(k)) for k half the size of m. Written iteratively with some bit shifting for division by 2, this should give you the theoretical O(log n) speed of exponentiation by squaring while staying entirely within integer arithmetic. (Well, O(log n) arithmetic operations. Since you will be working with numbers with roughly n bits, it won't be O(log n) time once you are forced to switch to a large integer library. After F(50), you will overflow the integer data type, which only goes up to 2^(31).)

(Apologies for not remembering Java well enough to implement this in Java; anyone who wants to is free to edit it in.)

Answer 4

Dynamic programming

Idea:Instead of recomputing the same value multiple times you just store the value calculated and use them as you go along.

f(n)=f(n-1)+f(n-2) with f(0)=0,f(1)=1. So at the point when you have calculated f(n-1) you can easily calculate f(n) if you store the values of f(n) and f(n-1).

Let's take an array of Bignums first. A[1..200]. Initialize them to -1.

Pseudocode

fact(n)
{
if(A[n]!=-1) return A[n];
A[0]=0;
A[1]=1;
for i=2 to n
  A[i]= addition of A[i],A[i-1];
return A[n]
}

This runs in O(n) time. Check it out yourself.

This technique is also called memoization.

The IDEA

Dynamic programming (usually referred to as DP ) is a very powerful technique to solve a particular class of problems. It demands very elegant formulation of the approach and simple thinking and the coding part is very easy. The idea is very simple, If you have solved a problem with the given input, then save the result for future reference, so as to avoid solving the same problem again.. shortly 'Remember your Past'.

If the given problem can be broken up in to smaller sub-problems and these smaller subproblems are in turn divided in to still-smaller ones, and in this process, if you observe some over-lappping subproblems, then its a big hint for DP. Also, the optimal solutions to the subproblems contribute to the optimal solution of the given problem ( referred to as the Optimal Substructure Property ).

There are two ways of doing this.

1.) Top-Down : Start solving the given problem by breaking it down. If you see that the problem has been solved already, then just return the saved answer. If it has not been solved, solve it and save the answer. This is usually easy to think of and very intuitive. This is referred to as Memoization. (I have used this idea).

2.) Bottom-Up : Analyze the problem and see the order in which the sub-problems are solved and start solving from the trivial subproblem, up towards the given problem. In this process, it is guaranteed that the subproblems are solved before solving the problem. This is referred to as Dynamic Programming. (MinecraftShamrock used this idea)

---

There's more!

(Other ways to do this)

Look our quest to get a better solution doesn't end here. You will see a different approach-

If you know how to solve recurrence relation then you will find a solution to this relation

f(n)=f(n-1)+f(n-2) given f(0)=0,f(1)=1

You will arrive at the formula after solving it-

f(n)= (1/sqrt(5))((1+sqrt(5))/2)^n - (1/sqrt(5))((1-sqrt(5))/2)^n

which can be written in more compact form

f(n)=floor((((1+sqrt(5))/2)^n) /sqrt(5) + 1/2)

Complexity

You can get the power a number in O(logn) operations. You have to learn the Exponentiation by squaring.

EDIT: It is good to point out that this doesn't necessarily mean that the fibonacci number can be found in O(logn). Actually the number of digits we need to calculate frows linearly. Probably because of the position where I stated that it seems to claim the wrong idea that factorial of a number can be calculated in O(logn) time. [Bakurui,MinecraftShamrock commented on this]

Answer 5

Precompute a large number of fib(n) results, and store them as a lookup table inside your algorithm. Bam, free "speed"

Now if you need to compute fib(101) and you already have fibs 0 to 100 stored, this is just like trying to compute fib(1).

Chances are this isn't what this homework is looking for, but it's a completely legit strategy and basically the idea of caching extracted further away from running the algorithm. If you know you're likely to be computing the first 100 fibs often and you need to do it really really fast, there's nothing faster than O(1). So compute those values entirely out of band and store them so they can be looked up later.

Of course, cache values as you compute them too :) Duplicated computation is waste.

Answer 6

Having followed a similar approach some time ago, I've just realized there's another optimization you can make.

If you know two large consecutive answers, you can use this as a starting point. For example, if you know F(100) and F(101), then calculating F(104) is approximately as difficult (*) as calculating F(4) based on F(0) and F(1).

Calculating iteratively up is as efficient calculation-wise as doing the same using cached-recursion, but uses less memory.

Having done some sums, I have also realized that, for any given z < n:

F(n)=F(z) * F(n-z) + F(z-1) * F(n-z-1)

If n is odd, and you choose z=(n+1)/2, then this is reduced to

F(n)=F(z)^2+F(z-1)^2

It seems to me that you should be able to use this by a method I have yet to find, that you should be able use the above info to find F(n) in the number of operations equal to:

the number of bits in n doublings (as per above) + the number of 1 bits in n addings; in the case of 104, this would be (7 bits, 3 '1' bits) = 14 multiplications (squarings), 10 additions.

(*) assuming adding two numbers takes the same time, irrelevant of the size of the two numbers.