Fibonacci Numbers: Difference between revisions

From NovaOrdis Knowledge Base
Jump to navigation Jump to search
 
(6 intermediate revisions by the same user not shown)
Line 2: Line 2:


* [[Mathematics]]
* [[Mathematics]]
* [[Algorithms#Dynamic_Programming_Algorithms|Algorithms]]
* [[Dynamic_Programming#Canonical_Use|Dynamic Programming]]


=Overview=
=Overview=
Line 13: Line 15:
F<sub>i</sub> = F<sub>i-1</sub> + F<sub>i-2</sub> for i ≥ 2.
F<sub>i</sub> = F<sub>i-1</sub> + F<sub>i-2</sub> for i ≥ 2.


<font color=darkgray>TODO [[CLRS]] page 108.</font>
<font color=darkkhaki>TODO [[CLRS]] page 108.</font>


=Golden Ratio=
=Golden Ratio=
Line 40: Line 42:
Running time < 2<sup>0</sup> + 2<sup>1</sup> + ... + 2<sup>n-1</sup>.
Running time < 2<sup>0</sup> + 2<sup>1</sup> + ... + 2<sup>n-1</sup>.


This is a bad exponential time, and attempting to use the algorithm for anything larger than 50 gets problematic. The solution is to apply a [[#Dynamic_Programming|dynamic programming]] method:
This is a bad exponential time, and attempting to use the algorithm for anything larger than 50 gets problematic. The fact that we're redundantly computing the same values again and again provides a hint that we could use a [[#Dynamic_Programming|dynamic programming]] method:


==Dynamic Programming==
==Dynamic Programming==
<syntaxhighlight lang='java'>
public static long fib_dp(int n) {
  long[] a = new long[n + 1];
  a[0] = 0;
  a[1] = 1;
  for(int i = 2; i <= n; i ++) {
    a[i] = a[i-1] + a[i-2];
  }
  return a[n];
}
</syntaxhighlight>
Also see: {{Algorithms#Dynamic_Programming_Algorithms|Dynamic Programming Algorithms}}

Latest revision as of 21:53, 11 November 2021

Internal

Overview

We define Fibonacci numbers by the following recurrence:

F0 = 0

F1 = 1

Fi = Fi-1 + Fi-2 for i ≥ 2.

TODO CLRS page 108.

Golden Ratio

φ=1.6180339...


Algorithms

Straightforward Recursive

A straightforward recursive algorithm looks like this:

public static long fib(long n) {
  if (n == 0) {
    return 0;
  }
  if (n == 1) {
    return 1;
  }
  return fib(n - 1) + fib(n - 2);
}

However, the running time of this method, computed with a recursion tree, is O(2n):

Fibonacci Recursion Tree.png

Running time < 20 + 21 + ... + 2n-1.

This is a bad exponential time, and attempting to use the algorithm for anything larger than 50 gets problematic. The fact that we're redundantly computing the same values again and again provides a hint that we could use a dynamic programming method:

Dynamic Programming

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

Also see: Template:Algorithms