Different ways to calculate fibonacci numbers15 Mar 2014
NOTE: All code below are written in a dialect of Scheme called Racket.
Problem: exponential complexity, even can’t calculate (fibonacci 1000).
This version is not only faster, but also takes constant space, as the tail recursion can be optimized by compiler.
Problem: if (fibonacci 1000) is called twice, it performs the same computation twice!
Idea: Use a table to remember computation resutls.
Idea: A faster approach to compute is to compute , which only requires 4 steps of computation. If the power n is large, it can reduce the complexity logarithmically.
In the fibonacci-iter above, the operation a’ = b and b’ = a + b is performed n times, which is equivalent to multiply the matrix n times. This fact allows us to perform the same optimization as in computing exponentiation.
You can verify that computes the n-th and (n+1)-th fibonacci numbers. The multiplier vector just selects the element c and d from the result matrix , so it can be omitted, as whenever we know the matrix, we know the result.
Another useful fact is that in the result matrix we have b = c and d = a + b. So the matrix can be written as .
Also note that according to matrix multiplication, following formula holds:
where and .
From the mathematical facts above, the implementation comes naturally.
The version above works, but we can do better to use tail recursion.
The improved version utilizes another mathematical fact:
where and .
It’s also helpful to know that for any integer n and matrix X:
where is either 0 or 1.
The formula above means that is the binary representation of n. The improved version uses p and q to accumulate the result of the first i terms, a and b remembers . If n can be divided by 2 for i times, it means the i-th position of n(i.e. ) is 0, then i can be increased, accumulator unchanged. Otherwise, it means the i-th position of n(i.e. ) is 1, then the accumulator p and q is updated, and i will be increased in the next recursion call as we’ve changed the last bit in to 0 through n - 1. The process stops if n is equal to zero, which means for every i bigger than its current value, is zero.
The complexity of this fast method is , compared to of the second approach above.
- SCIP section 1.2 and 3.3