This is a somewhat tricky problem and the code below may require some careful study to understand! You are quite right that hashes are the key, but even with that some thought is required. Note that the following code tries to keep the clusters in order of the first cluster entry found which complicates things a little more, but not much.

use strict;
use warnings;

my %idLookup;
my %firstLines;

while (<DATA>) {
    chomp;
    next unless length;

    my $pair = $_;
    my ($first, $second) = map {s/.*\.//; $_} split ' ';

    push @{$idLookup{$first}}, [$second, $pair];
    push @{$idLookup{$second}}, [$first, $pair];
    $firstLines{$first} = $. unless exists $firstLines{$first};
    $firstLines{$second} = $. + 0.5 unless exists $firstLines{$second}
+;
}

my @lineOrder = sort {$firstLines{$a} <=> $firstLines{$b}} keys %first
+Lines;
my $clusterNum;

for my $id (@lineOrder) {
    next unless exists $firstLines{$id};

    my %hits = ($id => undef);
    my @cluster;

    while (keys %hits) {
        my $hit = (keys %hits)[0];
        my $list = $idLookup{$hit};

        delete $hits{$hit};

        next unless exists $firstLines{$hit};
        delete $firstLines{$hit};

        for my $entry (@$list) {
            my ($entryId, $pair) = @$entry;

            next unless exists $firstLines{$entryId};
            push @cluster, $pair;
            $hits{$entryId} = 1;
        }
    }

    print "cluster", ++$clusterNum, "\n";
    print join "\n", @cluster, "\n";
}


__DATA__
ID5141.C1665 ID5141.C2448
ID5141.C1253 ID5144.C2039
ID5141.C1596 ID5144.C1956
ID5141.C1906 ID5144.C2149
ID5141.C1221 ID5144.C1956
ID5141.C2149 ID5141.C2386
ID5141.C2039 ID5142.C1221
ID5141.C5887 ID5141.C7685
ID5141.C1005 ID5142.C2808
ID5141.C1046 ID5141.C1596
ID5141.C2386 ID5141.C4990
ID5141.C7685 ID5141.C4888

[download]

Prints:

cluster1
ID5141.C1665 ID5141.C2448

cluster2
ID5141.C1253 ID5144.C2039
ID5141.C2039 ID5142.C1221
ID5141.C1221 ID5144.C1956
ID5141.C1596 ID5144.C1956
ID5141.C1046 ID5141.C1596

cluster3
ID5141.C1906 ID5144.C2149
ID5141.C2149 ID5141.C2386
ID5141.C2386 ID5141.C4990

cluster4
ID5141.C5887 ID5141.C7685
ID5141.C7685 ID5141.C4888

cluster5
ID5141.C1005 ID5142.C2808

[download]
Perl's payment curve coincides with its learning curve.

In reply to Re: clustering pairs by GrandFather
in thread clustering pairs by sugar

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