Collatz Conjecture - Iteration rather than Recursion - perl

Question

I am working on something for learning purposes where I have tackled Collatz using recursion. If you see below I make use of #_ and $_ to keep the for alive.
#!/usr/bin/env perl
sub collatz {
my ($num, $count) = #_;
$count++;
if ($num == 1) {
return $count;
} elsif ($num % 2 == 0) {
return collatz($num/2, $count);
} else {
return collatz($num*3 + 1, $count);
}
}
my $max = 0;
my $saved = 0;
for (1..1000) {
my $length = collatz($_, 0);
print "Num: " . $_ . " Length: " . $length . "\n";
if ($length > $max) {
$max = $length;
$saved = $_;
}
}
print "The longest sequence starts with " . $saved . "\n";
I am trying to use iteration instead of recursion but I just can't think of how to tackle this. I am not after the code in the question, I just want some tips / hints on how to tackle this to get the same result.
I suspect I will need to use a while or an until field.
Any help would be appreciated, again I don't want the exact answer.
Update
Here is my second attempt, which is giving me an error of
Can't return outside a subroutine at answer2.pl line 38.
my $number = 0;
my $counter = 0;
while ($number != 1000) {
$counter++;
if ($number == 1) {
return $counter;
}
elsif ($number % 2 == 0) {
return ($number / 2, $counter);
}
else {
return ($number * 3 + 1, $counter);
}
$number++;
}
print "number" . $number . "counter" . $counter . "\n";

Answer 1

Basically you have tail recursion, which is nice and simple to eliminate.
Instead of collatz calling itself to generate the next step in the sequence, you simply change the variables in-place and loop back to the top.
In its crudest form this would be
sub collatz2 {
my ($num, $count) = #_;
NEXT:
$count++;
if ($num == 1) {
return $count;
}
elsif ($num % 2 == 0) {
$num = $num / 2;
}
else {
$num = $num * 3 + 1;
}
goto NEXT;
}
but it should be written much more nicely than that.
I ended up with this
sub collatz {
my ($num) = #_;
my $count = 1;
while ($num > 1) {
$num = $num % 2 ? $num * 3 + 1 : $num / 2;
++$count;
}
$count;
}

Answer 2

Consider adding the logic that returns when the condition is met in the while.
Spoiler:
my $iter = 0;
while($num != 1){ #do stuff; $iter++ }

Answer 3

Just use a for or while loop with the end condition that your number == 1.
Spoiler:
use strict;
use warnings;
my $max_num = 0;
my $max_steps = 0;
for my $num (1..1000) {
my $steps = 0;
for (my $i = $num; $i != 1; $steps++) {
$i = $i % 2 ? 3 * $i + 1 : $i / 2;
}
print "Num: " . $num . " Length: " . $steps . "\n";
if ($steps > $max_steps) {
$max_num = $num;
$max_steps = $steps;
}
}
print "The longest sequence starts with " . $max_num . "\n";