I have written the following subroutine to find minimum, maximum and average at the same time:

sub min_max_avg {
    my $ref = shift;
    my ( $min, $max, $agv, $i );
    my ( $current_min, $current_max );
    if ( @{$ref} % 2 == 0 ) {
        ( $min, $max ) =
          $ref->[0] < $ref->[1]
          ? ( $ref->[0], $ref->[1] )
          : ( $ref->[1], $ref->[0] );
        $agv = $ref->[0] + $ref->[1];
        $i   = 2;
    }
    else {
        $min = $max = $agv = $ref->[0];
        $i = 1;
    }
    while ( $i < @{$ref} ) {
        ( $current_min, $current_max ) =
          $ref->[$i] < $ref->[ $i + 1 ]
          ? ( $ref->[$i], $ref->[ $i + 1 ] )
          : ( $ref->[ $i + 1 ], $ref->[$i] );
        $min = $current_min if ( $current_min < $min );
        $max = $current_max if ( $current_max > $max );
        $agv += $ref->[$i] + $ref->[ $i + 1 ];
        $i   += 2;
    }
    return ( $min, $max, $agv / @{$ref} );
}
[download]

Note that:

• From well known "Introduction to algorithms, 2nd Edition" (ISBN 0070131511), page 162:

  ...at most 3(n/2) comparisons are sufficient to find both the minimum and the maximum. The strategy is to maintain the minimum and maximum elements seen thus far. Rather than processing each element of the input by comparing it against the current minimum and maximum, at a cost of 2 comparisons per element, we process elements in pairs. We compare pairs of elements from the input first with each other, and then we compare the smaller to the current minimum and the larger to the current maximum, at a cost of 3 comparisons for every 2 elements.Setting up initial values for the current minimum and maximum depends on whether n is odd or even. If n is odd, we set both the minimum and maximum to the value of the first element, and then we process the rest of the elements in pairs. If n is even, we perform 1 comparison on the first 2 elements to determine the initial values of the minimum and maximum, and then process the rest of the elements in pairs as in the case for odd n. Let us analyze the total number of comparisons. If n is odd, then we perform 3n/2 comparisons. If n is even, we perform 1 initial comparison followed by 3(n - 2)/2 comparisons, for a total of 3(n/2)-2. Thus, in either case, the total number of comparisons is at most 3(n/2).

  Therefore, regarding the number of comparisons the code would have to work fine.

• The subroutine don't check the input parameters. Infact, it fails with an input like this:
  

  [qw//]
  [download]

  
  or
  

  [qw/foo bar/]
  [download]

Can someone suggest improvements to me?
Can some step of the code be written in a better way? In a more perlish way?
Consider that i would like to preserve the structure of the algorithm... and thanks to all in advance.

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