DISCARD_A => \&callback2,
            DISCARD_B => \&callback3,
            CHANGE    => \&callback4,
        },
        \&key_generator,
        @extra_args,
    );


=head1 INTRODUCTION

(by Mark-Jason Dominus)

I once read an article written by the authors of C<diff>; they said
that they worked very hard on the algorithm until they found the
right one.

I think what they ended up using (and I hope someone will correct me,
because I am not very confident about this) was the `longest common
subsequence' method.  In the LCS problem, you have two sequences of
items:

    a b c d f g h j q z

    a b c d e f g i j k r x y z

and you want to find the longest sequence of items that is present in
both original sequences in the same order.  That is, you want to find
a new sequence I<S> which can be obtained from the first sequence by
deleting some items, and from the secend sequence by deleting other
items.  You also want I<S> to be as long as possible.  In this case I<S>
is

    a b c d f g j z

From there it's only a small step to get diff-like output:

    e   h i   k   q r x y
    +   - +   +   - + + +

This module solves the LCS problem.  It also includes a canned function
to generate C<diff>-like output.

It might seem from the example above that the LCS of two sequences is
always pretty obvious, but that's not always the case, especially when
the two sequences have many repeated elements.  For example, consider

    a x b y c z p d q
    a b c a x b y c z

A naive approach might start by matching up the C<a> and C<b> that
appear at the beginning of each sequence, like this:

    a x b y c         z p d q
    a   b   c a b y c z

This finds the common subsequence C<a b c z>.  But actually, the LCS
is C<a x b y c z>:

          a x b y c z p d q
    a b c a x b y c z

or

    a       x b y c z p d q
    a b c a x b y c z

=head1 USAGE

(See also the README file and several example
scripts include with this module.)

This module now provides an object-oriented interface that uses less
memory and is easier to use than most of the previous procedural
interfaces.  It also still provides several exportable functions.  We'll
deal with these in ascending order of difficulty:  C<LCS>,
C<LCS_length>, C<LCSidx>, OO interface, C<prepare>, C<diff>, C<sdiff>,
C<traverse_sequences>, and C<traverse_balanced>.

=head2 C<LCS>

Given references to two lists of items, LCS returns an array containing
their longest common subsequence.  In scalar context, it returns a
reference to such a list.

    @lcs    = LCS( \@seq1, \@seq2 );
    $lcsref = LCS( \@seq1, \@seq2 );

C<LCS> may be passed an optional third parameter; this is a CODE
reference to a key generation function.  See L</KEY GENERATION
FUNCTIONS>.

    @lcs    = LCS( \@seq1, \@seq2, \&keyGen, @args );
    $lcsref = LCS( \@seq1, \@seq2, \&keyGen, @args );

Additional parameters, if any, will be passed to the key generation
routine.

=head2 C<LCS_length>

This is just like C<LCS> except it only returns the length of the
longest common subsequence.  This provides a performance gain of about
9% compared to C<LCS>.

=head2 C<LCSidx>

Like C<LCS> except it returns references to two arrays.  The first array
contains the indices into @seq1 where the LCS items are located.  The
second array contains the indices into @seq2 where the LCS items are located.

Therefore, the following three lists will contain the same values:

    my( $idx1, $idx2 ) = LCSidx( \@seq1, \@seq2 );
    my @list1 = @seq1[ @$idx1 ];
    my @list2 = @seq2[ @$idx2 ];
    my @list3 = LCS( \@seq1, \@seq2 );

=head2 C<new>

    $diff = Algorithm::Diffs->new( \@seq1, \@seq2 );
    $diff = Algorithm::Diffs->new( \@seq1, \@seq2, \%opts );

C<new> computes the smallest set of additions and deletions necessary
to turn the first sequence into the second and compactly records them
in the object.

You use the object to iterate over I<hunks>, where each hunk represents
a contiguous section of items which should be added, deleted, replaced,
or left unchanged.

=over 4

The following summary of all of the methods looks a lot like Perl code
but some of the symbols have different meanings:

    [ ]     Encloses optional arguments
    :       Is followed by the default value for an optional argument
    |       Separates alternate return results

Method summary:

    $obj        = Algorithm::Diff->new( \@seq1, \@seq2, [ \%opts ] );
    $pos        = $obj->Next(  [ $count : 1 ] );
    $revPos     = $obj->Prev(  [ $count : 1 ] );
    $obj        = $obj->Reset( [ $pos : 0 ] );
    $copy       = $obj->Copy(  [ $pos, [ $newBase ] ] );
    $oldBase    = $obj->Base(  [ $newBase ] );

Note that all of the following methods C<die> if used on an object that
is "reset" (not currently pointing at any hunk).

    $bits       = $obj->Diff(  );
    @items|$cnt = $obj->Same(  );
    @items|$cnt = $obj->Items( $seqNum );
    @idxs |$cnt = $obj->Range( $seqNum, [ $base ] );
    $minIdx     = $obj->Min(   $seqNum, [ $base ] );
    $maxIdx     = $obj->Max(   $seqNum, [ $base ] );
    @values     = $obj->Get(   @names );

Passing in C<undef> for an optional argument is always treated the same
as if no argument were passed in.

=item C<Next>

    $pos = $diff->Next();    # Move forward 1 hunk
    $pos = $diff->Next( 2 ); # Move forward 2 hunks
    $pos = $diff->Next(-5);  # Move backward 5 hunks

C<Next> moves the object to point at the next hunk.  The object starts
out "reset", which means it isn't pointing at any hunk.  If the object
is reset, then C<Next()> moves to the first hunk.

C<Next> returns a true value iff the move didn't go past the last hunk.
So C<Next(0)> will return true iff the object is not reset.

Actually, C<Next> returns the object's new position, which is a number
between 1 and the number of hunks (inclusive), or returns a false value.

=item C<Prev>

C<Prev($N)> is almost identical to C<Next(-$N)>; it moves to the $Nth
previous hunk.  On a 'reset' object, C<Prev()> [and C<Next(-1)>] move
to the last hunk.

The position returned by C<Prev> is relative to the I<end> of the
hunks; -1 for the last hunk, -2 for the second-to-last, etc.

=item C<Reset>

    $diff->Reset();     # Reset the object's position
    $diff->Reset($pos); # Move to the specified hunk
    $diff->Reset(1);    # Move to the first hunk
    $diff->Reset(-1);   # Move to the last hunk

C<Reset> returns the object, so, for example, you could use
C<< $diff->Reset()->Next(-1) >> to get the number of hunks.

=item C<Copy>

    $copy = $diff->Copy( $newPos, $newBase );

C<Copy> returns a copy of the object.  The copy and the orignal object
share most of their data, so making copies takes very little memory.
The copy maintains its own position (separate from the original), which
is the main purpose of copies.  It also maintains its own base.

By default, the copy's position starts out the same as the original
object's position.  But C<Copy> takes an optional first argument to set the
new position, so the following three snippets are equivalent:

    $copy = $diff->Copy($pos);

    $copy = $diff->Copy();
    $copy->Reset($pos);

    $copy = $diff->Copy()->Reset($pos);

C<Copy> takes an optional second argument to set the base for
the copy.  If you wish to change the base of the copy but leave
the position the same as in the original, here are two
equivalent ways:

    $copy = $diff->Copy();
    $copy->Base( 0 );

    $copy = $diff->Copy(undef,0);

Here are two equivalent way to get a "reset" copy:

    $copy = $diff->Copy(0);

    $copy = $diff->Copy()->Reset();

=item C<Diff>

    $bits = $obj->Diff();

C<Diff> returns a true value iff the current hunk contains items that are
different between the two sequences.  It actually returns one of the
follow 4 values:

=over 4

=item 3

C<3==(1|2)>.  This hunk contains items from @seq1 and the items
from @seq2 that should replace them.  Both sequence 1 and 2
contain changed items so both the 1 and 2 bits are set.

=item 2

This hunk only contains items from @seq2 that should be inserted (not
items from @seq1).  Only sequence 2 contains changed items so only the 2
bit is set.

=item 1

This hunk only contains items from @seq1 that should be deleted (not
items from @seq2).  Only sequence 1 contains changed items so only the 1
bit is set.

=item 0

This means that the items in this hunk are the same in both sequences.
Neither sequence 1 nor 2 contain changed items so neither the 1 nor the
2 bits are set.

=back

=item C<Same>

C<Same> returns a true value iff the current hunk contains items that
are the same in both sequences.  It actually returns the list of items
if they are the same or an emty list if they aren't.  In a scalar
context, it returns the size of the list.

=item C<Items>

    $count = $diff->Items(2);
    @items = $diff->Items($seqNum);

C<Items> returns the (number of) items from the specified sequence that
are part of the current hunk.

If the current hunk contains only insertions, then
C<< $diff->Items(1) >> will return an empty list (0 in a scalar conext).
If the current hunk contains only deletions, then C<< $diff->Items(2) >>
will return an empty list (0 in a scalar conext).

If the hunk contains replacements, then both C<< $diff->Items(1) >> and
C<< $diff->Items(2) >> will return different, non-empty lists.

Otherwise, the hunk contains identical items and all of the following
will return the same lists:

    @items = $diff->Items(1);
    @items = $diff->Items(2);
    @items = $diff->Same();

=item C<Range>

    $count = $diff->Range( $seqNum );
    @indices = $diff->Range( $seqNum );
    @indices = $diff->Range( $seqNum, $base );

C<Range> is like C<Items> except that it returns a list of I<indices> to
the items rather than the items themselves.  By default, the index of
the first item (in each sequence) is 0 but this can be changed by
calling the C<Base> method.  So, by default, the following two snippets
return the same lists:

    @list = $diff->Items(2);
    @list = @seq2[ $diff->Range(2) ];

You can also specify the base to use as the second argument.  So the
following two snippets I<always> return the same lists:

    @list = $diff->Items(1);
    @list = @seq1[ $diff->Range(1,0) ];

=item C<Base>

    $curBase = $diff->Base();
    $oldBase = $diff->Base($newBase);

C<Base> sets and/or returns the current base (usually 0 or 1) that is
used when you request range information.  The base defaults to 0 so
that range information is returned as array indices.  You can set the
base to 1 if you want to report traditional line numbers instead.

=item C<Min>

    $min1 = $diff->Min(1);
    $min = $diff->Min( $seqNum, $base );

C<Min> returns the first value that C<Range> would return (given the
same arguments) or returns C<undef> if C<Range> would return an empty
list.

=item C<Max>

C<Max> returns the last value that C<Range> would return or C<undef>.

=item C<Get>

    ( $n, $x, $r ) = $diff->Get(qw( min1 max1 range1 ));
    @values = $diff->Get(qw( 0min2 1max2 range2 same base ));

C<Get> returns one or more scalar values.  You pass in a list of the
names of the values you want returned.  Each name must match one of the
following regexes:

    /^(-?\d+)?(min|max)[12]$/i
    /^(range[12]|same|diff|base)$/i

The 1 or 2 after a name says which sequence you want the information
for (and where allowed, it is required).  The optional number before
"min" or "max" is the base to use.  So the following equalities hold:

    $diff->Get('min1') == $diff->Min(1)
    $diff->Get('0min2') == $diff->Min(2,0)

Using C<Get> in a scalar context when you've passed in more than one
name is a fatal error (C<die> is called).

=back

=head2 C<prepare>

Given a reference to a list of items, C<prepare> returns a reference
to a hash which can be used when comparing this sequence to other
sequences with C<LCS> or C<LCS_length>.

    $prep = prepare( \@seq1 );
    for $i ( 0 .. 10_000 )
    {
        @lcs = LCS( $prep, $seq[$i] );
        # do something useful with @lcs
    }

C<prepare> may be passed an optional third parameter; this is a CODE
reference to a key generation function.  See L</KEY GENERATION
FUNCTIONS>.

    $prep = prepare( \@seq1, \&keyGen );
    for $i ( 0 .. 10_000 )
    {
        @lcs = LCS( $seq[$i], $prep, \&keyGen );
        # do something useful with @lcs
    }

Using C<prepare> provides a performance gain of about 50% when calling LCS
many times compared with not preparing.

=head2 C<diff>

    @diffs     = diff( \@seq1, \@seq2 );
    $diffs_ref = diff( \@seq1, \@seq2 );

C<diff> computes the smallest set of additions and deletions necessary
to turn the first sequence into the second, and returns a description
of these changes.  The description is a list of I<hunks>; each hunk
represents a contiguous section of items which should be added,
deleted, or replaced.  (Hunks containing unchanged items are not
included.)

The return value of C<diff> is a list of hunks, or, in scalar context, a
reference to such a list.  If there are no differences, the list will be
empty.

Here is an example.  Calling C<diff> for the following two sequences:

    a b c e h j l m n p
    b c d e f j k l m r s t

would produce the following list:

    (
      [ [ '-', 0, 'a' ] ],

      [ [ '+', 2, 'd' ] ],

      [ [ '-', 4, 'h' ],
        [ '+', 4, 'f' ] ],

      [ [ '+', 6, 'k' ] ],

      [ [ '-',  8, 'n' ],
        [ '-',  9, 'p' ],