The old mind puzzler, now as a 90 byte function.
var dollar = 100; var coins = [100, 50, 25, 10, 5, 1]; f(dollar, coins); // 293
var twoPound = 200; var coins = [200, 100, 50, 20, 10, 5, 2, 1]; f(twoPound, coins); // 73682
U mad Project Euler? http://projecteuler.net/problem=31
| function f( | |
| a, // target count | |
| b, // array of currencies to use | |
| c, // placeholder for current count | |
| d, // placeholder for array index | |
| e // placeholder for counter | |
| ) { | |
| for (c |= e = 0; // initial: set counter(e) to zero. if current count(c) is undefined, set it to 0 to allow adding | |
| b[d |= 0] ? // conditional: if we haven't hit the end of the index | |
| c <= a : // evaluate whether we've gone over the limit | |
| (e = c == a, 0); // else set the count to zero or 1 and return false to stop the loop | |
| c += b[d] // increment: add currency to current count | |
| ) | |
| e += f(a, b, c, d + 1); // recursion. use the next item in the array and call the same function, add it to the count | |
| return e // return the count | |
| } |
| function(a,b,c){var d=i=0,e=function(b){d+=b==a};while(c=b[i++])(function(b,c){e=function(d,e){for(e=d;e<=a;e+=b)c(e)}})(c,e);return e(0),d} | |
| function f(a,b,c,d,e){for(c|=e=0;b[d|=0]?c<=a:(e=c==a,0);c+=b[d])e+=f(a,b,c,d+1);return e} |
| function f(a,b,c,d,e){for(c|=e=0;b[d|=0]?c<=a:(e=c==a,0);c+=b[d])e+=f(a,b,c,d+1);return e} |
| DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE | |
| Version 2, December 2004 | |
| Copyright (C) 2011 Tim Wood <timwoodcreates.com> | |
| 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": "currencyPermutations", | |
| "description": "A currency permutation calculator (aka Project Euler 31)", | |
| "keywords": [ | |
| "euler", | |
| "currency", | |
| "coins", | |
| "permutation" | |
| ] | |
| } |
| <!DOCTYPE html> | |
| <title>Currency Permutations</title> | |
| <div>Expected value: <b>293</b></div> | |
| <div>Actual value: <b id="ret"></b></div> | |
| <script> | |
| function f(a,b,c,d,e){for(c|=e=0;b[d|=0]?c<=a:(e=c==a,0);c+=b[d])e+=f(a,b,c,d+1);return e} | |
| document.getElementById( "ret" ).innerHTML = f(100, [100, 50, 25, 10, 5, 1]); | |
| </script> |
That's what I meant, but now re-reading it, I can see how it's unclear. I meant 'use c as c if already set'. I'll clean it up.
Hmm... I think you misunderstand the
c |= e = 0.It's actually
c = c | (e = 0), which means each function call owns a counter, and|0is to prevent the c as undefined in the first call.