Reputation: 19315
I'm in PHP working on an Euler problem. I have this function so far:
<?php
$biggest = 0;
$counter = 1;
function test($i){
global $biggest;
global $counter;
if ($i == 1) {
echo "I'm done! Took me $biggest steps";
}
else {
if ($i%2 == 0) {
$counter = $counter + 1;
if ($counter>$biggest) {
$biggest = $counter;
}
test($i/2);
}
else {
$counter = $counter + 1;
if ($counter>$biggest) {
$biggest = $counter;
}
test(3*$i+1);
}
}
}
test(13);
?>
I have the problem mostly licked, but I can't seem to get back at the original input. The question is "When you have a number, if odd get 3n+1, when even, get n/2, do until returns 1. What starting value yields the most "steps" before you get to one?" I currently am returning the number of steps, but I keep resetting $i as I recurse, so I can't record what starting # yielded my $biggest number of steps.
How can I keep track of that number, but also not have it destroyed at the next instance of the loop? (I'll eventually wrap this in a for ($i=1, $i<1000000, $i++) loop)
Thanks!
Upvotes: 2
Views: 1144
Reputation: 60150
A common approach is to pass the original argument through each time, so that when eventually you get to your base case, you still have it available. A primitive (and almost entirely unrelated example):
<?php
function fact($n) {
if($n == 1) return 1;
else return $n * fact($n - 1);
}
?>
This is an extremely basic implementation of the factorial function in PHP. Now say you wanted for whatever reason to have the initial value available in the final step: you'd build a wrapper function fact($n)
that would call something like memory_fact($n, $initial)
:
<?php
function fact($n) {
return memory_fact($n, $n);
}
function memory_fact($n, $initial) {
if($n == 1) return 1;
else return $n * memory_fact($n - 1, $initial);
}
?>
This way, memory_fact
always knows where it started.
Upvotes: 3
Reputation: 77450
You don't need the globals; globals are evil. Try returning something useful from test()
. Also, you'll find the test()
above wastes many cycles. Try using memoization.
Here's a memoization example for calculating Fibonacci numbers:
function fib($n) {
static $data = array(1, 1);
if (!isset($data[$n])) {
$data[$n] = fib($n-1) + fib($n-2);
}
return $data[$n];
}
Note that there are other time-efficent constant-space approaches to handle Fibonacci numbers (including one in O(log n) time), but the Collatz conjecture is a little trickier.
Upvotes: 0
Reputation: 258408
It's easy, just pass it around as a parameter! Here's some python-ish pseudocode:
def func(start, arg):
if foo(arg):
return func(start, arg+1)
else:
return [start, arg]
Upvotes: 0