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### Re^2: Finding the max()/min()

by Aristotle (Chancellor)
 on Dec 18, 2004 at 02:15 UTC Need Help??

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

Here's another recursive form:

```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.

Replies are listed 'Best First'.
Re^3: Finding the max()/min()
by elusion (Curate) on Dec 18, 2004 at 17:26 UTC
It's normally best to end recursive solutions with the recursive call. That allows the compiler to use tail-recursion to speed up the solution.

Of course, in Perl you need to jump through a few hoops to do use tail-recursion, 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 tail-recursion.

There's a bug in that code, try finding the max of this list...
```@a = (1,0,3,2);

-- All code is 100% tested and functional unless otherwise noted.
Good catch; I forgot the edge case. Here's the fix:
```sub max {
my \$max = shift;
return \$max if not @_;
my \$next = shift;
unshift @_, \$max > \$next ? \$max : \$next;
goto &max;
}

That's all well and nice, but my goal was to use an approach which directly generalizes from the specific case of a two element list. As you'll also notice I used the value returned from the recursive call twice, it's both kind of difficult and beside the point to work on tail recursion here. Noone in their right mind writes a recursive max() outside a functional language anyway, and in a pure functional language I'd just write max( @l ) twice and let the compiler/VM figure out that the result need only be calculated once.

In any case, your new approach is just a variation on what was already posted in the parent node — no need for repetition…

Makeshifts last the longest.

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