views:

137

answers:

3

I would like to implement a binary search algorithm in Perl. My 'array' is sorted in decreasing order (not an actual array, but a function that gets an index and returns values). the problem is that there might be stretches of identical values. If my searched value is in such a stretch, I want to return the first index that contains it.

This is what I have written:

# get_val should be a *decreasing* function for idexes $i in min..max,
# formally: for any $i,$j s.t. $max>=$i>$j>=$min :
# $get_val_subref($i, $extra) <= $get_val_subref($j, $extra)
# min and max are the inclusive boundaries for the search
# get_val sub should get an index in min..max and an extra data reference, and return
# the value for the given index
# returns the smallest index $i in min..max for which $get_val_subref($j, $extra)
# returns $searched_val, or undef if no such index exists
sub binary_search {
    my ( $min, $max, $searched_val, $get_val_subref, $get_val_sub_extra_data )
        = @_;
    my ( $mid, $val );
    while ( $min <= $max ) {
        $mid = $min + int( ( $max - $min ) / 2 );
        $val = $get_val_subref->( $mid, $get_val_sub_extra_data );

        if ( $val > $searched_val ) {
            $min = $mid + 1;
        }
        elsif ( $val < $searched_val ) {
            $max = $mid - 1;
        }
        else { ## SEE MY QUESTION BELOW ##

            # surely $val == $searched_val, but is it the first one?

            if (    $mid > $min
                and $get_val_subref->( $mid - 1, $get_val_sub_extra_data )
                == $searched_val )
            {

                # $val == $searched_val and prev($val) == $searched_val
                # we have to continue
                $max = $mid - 1;
            }
            else {

                # $val == $searched_val and prev($val) != $searched_val
                # wer'e done
                return $mid;
            }
        }
    }

    # $val was not found. return undef
    return undef;

}

and this is a simple example for using it:

sub get_val_sub {
    my ( $pos, $a ) = @_;
    my $val = $a->[$pos];
    return $val;
}

my @arr = (80, 40, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0);
 say "RET:", binary_search( 0, $#arr, 0, \&get_val_sub, \@arr );

The problem is I'm not sure my last else (marked with ## SEE MY QUESTION BELOW ##) is "pretty". Is there a better way of doing that?

+4  A: 

Although I initially agreed with Axeman's answer ... it is, in a small way, similar to my first (really bad) answer in using linear logic (at least a small bit of it). Specifically, there is no reason to call $get_val_subref with $mid - 1. That's an unnecessary linear search step.

Here's what I would suggest. In addition to avoiding linear searching, it has the benefit of being extremely simple:

sub binary_search {
    ...
    my ( $mid, $val, $solution );
    while ( $min <= $max ) {
        ...
        else {
            $solution = $mid; # Store a possible solution.
            $max = $mid - 1;  # But continue with the binary search
                              # until $min and $max converge on each other.
        }
    }
    return $solution;
}
FM
+1  A: 

Although I first agreed with FM's answer, the case that you show (with all the zeroes) is not a good case for a linear back search. And although I didn't like that you simply continued the binary search, "the first x" does have a computable value, and would still have sub-linear performance, while the linear back search has--of course--a linear one.

So I like your idea, but it's more compact like this:

else {  
    return $mid unless 
        (   $mid > $min
        and $get_val_subref->( $mid - 1, $get_val_sub_extra_data )
            == $searched_val
        );
    $max = $mid - 1;
}

The linear back search is an easier computation, but as the value functions get more complex, the fewer the computations the better.

Axeman
A: 

You may be looking for Newton's approximation method.

reinierpost