Aug. 5th, 2015

madbernard: a long angled pier (Default)
Factorial happened last week, also; the CoderByte challenge was, we'll test on numbers between 1 and 18, give us that number multiplied by every non-0 integer less than it. Ben-the-housemate suggested that this could be done recursively, so I tried that first, but didn't hack it; so, given the ticking clock, I aborted to the exact same decreasing for-loop plan I'd used last week. Looking through other answers was gratifying... I was among the most elegant! I found the recursive method from user mattlarsh, in a tiny tidy package:
function FirstFactorial(num) { 
  return num<=1?1:num*(FirstFactorial(num-1)); 
He's using the ternary operator there, the shortest possible way to say if/then. IF: the num is less than or equal to 1, DO THIS: return 1, AND IF IT'S NOT: multiply the num by FirstFactorial num-1. So once he has a stack of not-quite-resolved functions and the num gets to 1, it returns 1 and then the rest of the stack has all the numbers it needs to work with and can resolve the final return.

My first pass on recursion was the below, and I got stack overflow (infinite program), which I can see now is because the num kept getting bigger but the recursion would only end when the num got to 0.
function FirstFactorial(num) { 
if (num > 0) {
num *= (num - 1);
  return num; 


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