Using a hash is a good solution. I used the following algorithm: If both the numbers to be made related already belong to different classes, the classes are "merged", i.e. all numbers belonging to the class with a greater index are re-classified to the other class.

#!/usr/bin/perl
use warnings;
use strict;

use Data::Dumper;

my @a = (1, 1, 2, 3, 4, 4, 8, 8);
my @b = (3, 4, 3, 5, 6, 7, 9, 10);

print Dumper related(\@a, \@b);

sub related {
    my @a = @{ +shift };
    my @b = @{ +shift };
    die "Different length.\n" if @a != @b;

    my %r;
    PAIR: for my $i (0 .. $#a) {
        my @classes;
        for my $e ($a[$i], $b[$i]) {
            push @classes, $r{$e} if exists $r{$e};
        }

        # Both numbers already classified. Merge their classes if diff
+erent.
        if (@classes == 2) {
            next PAIR if $classes[0] == $classes[1];
            my ($min, $max) = sort { $a <=> $b } @r{$a[$i], $b[$i]};
            $r{$_} = $min for grep $r{$_} == $max, keys %r;

        # Just one number already classified. Classify the second one 
+to the same class.
        } elsif (@classes == 1) {
            if (exists $r{$a[$i]}) {
                $r{$b[$i]} = $r{$a[$i]};
            } else {
                $r{$a[$i]} = $r{$b[$i]};
            }

        # Both numbers are seen for the first time.
        } else { # @classes == 0
            my $min = $a[$i] < $b[$i] ? $a[$i] : $b[$i];
            @r{$a[$i], $b[$i]} = ($min) x 2;
        }
    }
    return \%r;
}
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