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

[Aristotle]

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 @_;
}

[elusion]

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.

[sleepingsquirrel]

There's a bug in that code, try finding the max of this list...

@a = (1,0,3,2);

[elusion]

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;
}

[Aristotle]

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…