为什么计算斐波纳契数列2 ^ n和不是n ^ 2的复杂性？ [英] Why is the complexity of computing the Fibonacci series 2^n and not n^2?

问题描述

我想用递归树来寻找斐波那契数列的复杂性，并得出结论树的高度= O（N）最坏的情况下，成本每个级别= CN ，所以的复杂性= N * N = N ^ 2

I am trying to find complexity of Fibonacci series using a recursion tree and concluded height of tree = O(n) worst case, cost of each level = cn, hence complexity = n*n=n^2

为什么是 O（2 ^ n）的？

推荐答案

在一个天真的递归斐波纳契的复杂性确实2ⁿ。

The complexity of a naive recursive fibonacci is indeed 2ⁿ.

T(n) = T(n-1) + T(n-2) = T(n-2) + T(n-3) + T(n-3) + T(n-4) = 
= T(n-3) + T(n-4) + T(n-4) + T(n-5) + T(n-4) + T(n-5) + T(n-5) + T(n-6) = ...

在你拨打每一步骤 T 的两倍，因此将提供最终的渐近障碍：
T（N）=2⋅2⋅...⋅2=2ⁿ

In each step you call T twice, thus will provide eventual asymptotic barrier of:
T(n) = 2⋅2⋅...⋅2 = 2ⁿ

奖金：最好的理论落实到斐波那契数实际上是一个密切公式，使用黄金比例：

bonus: The best theoretical implementation to fibonacci is actually a close formula, using the golden ratio:

Fib(n) = (φⁿ – (–φ)⁻ⁿ)/sqrt(5) [where φ is the golden ratio]

（但是，从现实生活中的precision错误遭受由于浮点算术，这是不准确）

(However, it suffers from precision errors in real life due to floating point arithmetics, which are not exact)

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