Thank you, David, for posting the transformation rules; that helped. jaredor provided an excellent solution for you. Here's an option that uses a rules-embedded s///eg on the two columns as strings:

use Modern::Perl;

my ( @table, $id1, $id2, $i );

push @table, [ (split) ] while <DATA>;

do { $id1 .= $_->[2]; $id2 .= $_->[3] } for @table;
$id1 =~ s/(?<=(.))(0+)(?=(.))/$1==$3?$1 x length $2:$2/eg;
$id2 =~ s/(?<=(.))(0+)(?=(.))/$1==$3?$1 x length $2:$2/eg;

$i = 0;
say "$_->[0]\t$_->[1]\t"
  . substr( $id1, $i,   1 ) . "\t"
  . substr( $id2, $i++, 1 )
  for @table;

__DATA__
1    100    1    2
101    200    1    2
201    300    1    0
301    400    1    0
401    500    0    2
501    600    0    2
601    700    0    0
701    800    1    1
801    900    1    2
[download]

Output:

1    100    1    2
101    200    1    2
201    300    1    2
301    400    1    2
401    500    1    2
501    600    1    2
601    700    1    0
701    800    1    1
801    900    1    2
[download]

This solution uses capture results, from within positive look-ahead and positive look-behind assertions, to check the two zero-enclosing digits for sameness by executing a ternary statement whose result is substituted for the enclosed zero(s).