in reply to Re^2: Finding the max()/min()
in thread Finding the max()/min()

Your solution is calling max(@xs) twice each step. This leads to O(2^n) growth for what should be an O(n) problem. Modifying it slightly to call max once and save the value in a temp variable should save considerable time on long lists.
sub max { my ($x, @xs) = @_; @xs ? do { my $m = maxdo(@xs); ($x, $m)[$x < $m] } : $x; }