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Created February 2, 2012 14:01
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Recursive Fibonacci with Caching
function f(n){
return f[n] = // Assign and return the cached n-th fibonacci number. Were using the function itself to cache the numbers - cool uh?
2>n ? // If we're looking for 0 or 1 ...
n : // ... just return it, else ...
f[n] || // ... look if we allready have the number in the cache or ...
f(n-1) + f(n-2) // ... calculate it recursively.
}
function f(n){return f[n]=2>n?n:f[n]||f(n-1)+f(n-2)}
DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE
Version 2, December 2004
Copyright (C) 2011 YOUR_NAME_HERE <YOUR_URL_HERE>
Everyone is permitted to copy and distribute verbatim or modified
copies of this license document, and changing it is allowed as long
as the name is changed.
DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE
TERMS AND CONDITIONS FOR COPYING, DISTRIBUTION AND MODIFICATION
0. You just DO WHAT THE FUCK YOU WANT TO.
{
"name": "fibonacciWithCache",
"description": "Recursive Fibonacci number calculation with caching.",
"keywords": [
"number",
"cache",
"recursive",
]
}
<!DOCTYPE html>
<title>Fibonacci Test</title>
<div>Expected value: <b>190392490709135</b></div>
<div>Actual value: <b id="ret"></b></div>
<script>
var myFunction = function f(n){return f[n]=2>n?n:f[n]||f(n-1)+f(n-2)}
document.getElementById("ret").innerHTML = myFunction(70)
</script>
@tsaniel
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tsaniel commented Feb 3, 2012

There is no need to use closure.
So just function f(n){return f[n]=2>n?n:f[n]||f(n-1)+f(n-2)} is ok.

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