Re: Finding the max()/min() by Zaxo (Archbishop) on Nov 11, 2004 at 02:38 UTC 
Here's a pair I like, similar to your delightfully symmeteric third example,
# max
my $max = ($x, $y)[$x < $y];
# min
my $min = ($x, $y)[$x > $y];
Those are slightly forthish, in lifting a logical value to arithmetic use.
Update: As subroutines,
sub max ($$) { $_[$_[0] < $_[1]] }
sub min ($$) { $_[$_[0] > $_[1]] }
I used prototypes there because the usual perl min and max extracts the extreme of a list, instead of from just two values.
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Re: Finding the max()/min() by elusion (Curate) on Nov 11, 2004 at 02:48 UTC 
What I really like is writing generic min/max routines that handle more than two variables. They show the beauty of recursive functions, even though Perl doesn't handle tailrecursion^{1}.
Iterative:
sub max {
my ($max, @vars) = @_;
for (@vars) {
$max = $_ if $_ > $max;
}
return $max;
}
Recursive:
sub max {
my ($max, $next, @vars) = @_;
return $max if not $next;
return max( $max > $next ? $max : $next, @vars );
}
Interestingly enough, though this example works well in several languages, Perl's iterative version is remarkably short; I've never noticed this before. I guess I just proved myself wrong.
^{1}Okay, so this isn't strictly true. Using a magic goto with the & sigil and @_, you can fake it. But it's not very pretty.
Update: Fixed my mistake. Thanks, pg. (I always want to use postmodifier if and for in the same statement.)
Update: Okay, so I should test my code before I post. Or find where I left my brain. Snippet 2 is fixed.  [reply] [d/l] [select] 

Update:
Per elusion's update, he has already fixed both solutions.
============
Your solution one does not compile.
Your solution two does not work, and it does not print anything:
use strict;
use warnings;
print max(1.23,2.6,55.1,4,5);
sub max {
my ($max, $next, @vars) = @_;
return if not $next;
return max( $max > $next ? $max : $next, @vars );
}
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Perhaps this will count as "cheating" for the purpose of this thread, but I just do this to find the min/max of a list
use List::Util qw(min max);
$min = min @list;
$max = max @list;
;)  [reply] [d/l] 

$min = reduce { ($a,$b)[$a < $b] } @list;
$min_key = reduce { ($a,$b)[$x{$a} < $x{$b}] } keys %x;
Working out max is left as an exercise :)
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sub max {
my $max=shift;
$max < $_ and $max = $_
for @vars;
return $max;
}
Is marginally shorted and probably faster, but commits what some would call the crime of logic based control flow.
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Your recursive version of max is broken for lists like (1,0,1,2) and (1,undef,1,2). i.e. the code assumes it is at the end of the list whenever $next==0 or $next==undef (which isn't true in general).
Here's a snazzy (if not the most efficient) recursive version with a hat tip to Zaxo...
sub max {
my ($x, @xs) = @_;
@xs ? ($x, max(@xs))[$x < max(@xs)] : $x
}
 All code is 100% tested and functional unless otherwise noted.
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Your solution is calling max(@xs) twice each step. This leads to O(2^n) growth for what should be an O(n) problem. Modifying it slightly to call max once and save the value in a temp variable should save considerable time on long lists.
sub max {
my ($x, @xs) = @_;
@xs ? do { my $m = maxdo(@xs); ($x, $m)[$x < $m] } : $x;
}
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sub max {
my( $i, @l ) = @_;
my $j = @l ? max( @l ) : $i;
return $i > $j ? $i : $j;
}
And for that matter, an iterative form:
sub max {
$_[ 0 ] < $_[ 1 ] ? shift : pop while @_ > 1;
return @_;
}
Makeshifts last the longest.
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It's normally best to end recursive solutions with the recursive call. That allows the compiler to use tailrecursion to speed up the solution.
Of course, in Perl you need to jump through a few hoops to do use tailrecursion, so I mentioned but didn't include it in my original node. But with a bit if tinkering you can use it, and here it is now:
sub max {
my $max = shift;
my $next = shift;
return $max if not $next;
unshift @_, $max > $next ? $max : $next;
goto &max;
}
Which I estimate to be about 10x faster on a random 5000 element list. (Exact benchmarks are left to the reader as an exercise :).
I wonder if Ponie (Perl 5 On New Internals Engine) will be able to detect and use tailrecursion.
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Re: Finding the max()/min() by pg (Canon) on Nov 11, 2004 at 03:25 UTC 
A recursive solution for a list of numbers: (the idea to get max for a list is originally gotten from elusion's post)
use strict;
use warnings;
print max(1.23,2.6,55.1,1.11,4,5);
print min(1.23,2.6,55.1,1.11,4,5);
sub max {
splice(@_, ($_[0] > $_[1]) ? 1 : 0, 1);
return ($#_ == 0) ? $_[0] : max(@_);
}
sub min {
splice(@_, ($_[0] > $_[1]) ? 0 : 1, 1);
return ($#_ == 0) ? $_[0] : min(@_);
}
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Re: Finding the max()/min() by monoxide (Beadle) on Nov 11, 2004 at 03:27 UTC 
A bit longer (just a little bit) but here goes. I realise that it can probably be shortened, but in the interests of brevity and my sanity here is the longer version.
max(3,4);
sub max {
($x,$y)=@_;@xd=split//,$x;
if(length($x) != length($y))
{
return $x if(length($x)>length($y));
return $y if(length($y)>length($x));
}
@yd=split//,$y;
for($i;$i<length($x);$i++)
{
if (ord($xd[$i])!= ord($yd[$i]))
{
return $y if( ord($y)>ord($x) );
return $y if(ord ($y) >ord($x ));
}
}
}
# *** UNTESTED ***
# It could be used to compare binary values,
# but i suppose that is outside the scope of
# this post.
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Re: Finding the max()/min() by etcshadow (Priest) on Nov 11, 2004 at 04:55 UTC 
Here's an interesting one that doesn involve a comparison operator (even an obfuscated one):
($x + $y + abs($x  $y)) / 2
The math behind that is: take the average of the two numbers, and then add half their difference (start halfway between, and then go up by half). So: (x+y)/2 + abs(xy)/2, and then factor out the division by two.

:Wq
Not an editor command: Wq
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Very cool. <nitpick>...but there's still a comparison; it's just hidden in the abs() function: (roughly:) abs(x) := x >= 0 ? x : x.</nitpick>
[OT] Reminds me of a programming assignment where we were supposed to put the square roots of the integers from 1100 inclusive into two groups with roughly equal sums. I used the fact that sqrt(x) + sqrt(x+3) is very close to sqrt(x+1) + sqrt(x+2) and managed to outperform all the other submissions by several orders of magnitude in both speed and "difference in sum". (Yea math!)
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Re: Finding the max()/min() by tmoertel (Chaplain) on Nov 11, 2004 at 05:12 UTC 
As others have mentioned, it's handy to capture min and
max logic in subroutines for reuse. Most times, if you have a
max subroutine, for example, it will take an array and return
the maximum element. That's great if you want to hold all of your
values in memory, but what if you don't? What if you want to compute
your maximum iteratively, e.g., one number at a time while reading
from a file?
As I offered in Re: getting the highest value in a simpler way, we can encapsulate our logic inside
of a closurebased factory function that makes "maximum finders":
sub make_max_finder {
my $max;
sub {
for (@_) { $max = $_ if !defined $max  $_ > $max }
$max;
}
}
The nice thing about this approach is that the logic is captured in
one place and yet we can use it both on big arrays and upon
iteratedover data. The array method:
make_max_finder>(1..10); # 10
The iterative, lightonmemory method:
my $max_finder = make_max_finder();
$max_finder>($_) while <>;
$max_finder>(); # retrieve maximum
Cheers, Tom
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Re: Finding the max()/min() by brian_d_foy (Abbot) on Nov 11, 2004 at 06:01 UTC 
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Is List::Util now part of the core? It wasn't in 5.6, which made me not use it.
Being right, does not endow the right to be rude; politeness costs nothing. Being unknowing, is not the same as being stupid. Expressing a contrary opinion, whether to the individual or the group, is more often a sign of deeper thought than of cantankerous belligerence. Do not mistake your goals as the only goals; your opinion as the only opinion; your confidence as correctness. Saying you know better is not the same as explaining you know better.
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Re: Finding the max()/min() by Zed_Lopez (Chaplain) on Nov 11, 2004 at 17:49 UTC 
sub min { eigenstates( any(@_) <= all(@_) ) }
sub max { eigenstates( any(@_) >= all(@_) ) }
In the real world, I'd use List::Util::max; if I wrote my own, it'd be this small variation on solutions above:
sub max {
my $max = shift;
$max = $max > $_ ? $max : $_ for @_;
return $max
}
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Re: Finding the max()/min() by cLive ;) (Parson) on Nov 11, 2004 at 18:09 UTC 
$max = $x%$y$x ? $x : $y
cLive ;)
update: d'oh. caveat  "for $x and $y both +ve integers lol...  [reply] [d/l] 

$x=1; $y=0;
$x=1; $y=2;
$x=1.2; $y=2;
 All code is 100% tested and functional unless otherwise noted.
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