Here's a simplistic approach using regexps:

  my @a = (
    [2,5,10,5,12,6,21,5,10,12,23],
    [5,6,11,10,5,10,6,21,5,1,9],
    [6,5,10,15,21]
  );
  my($m, $n) = (2, 2);
  my $joined = join ';', map join(',', @$_), @a;

  while ($joined =~ /
    \b
    (?=
      ( @{[ join ',', ('\d+') x $n ]} )
      \b
      @{[ '(?> .*? \b \1 \b )' x ($m - 1) ]}
    )
  /xg) {
    print "($1)\n" unless $seen{$1}++;
  }

[download]

I suspect this isn't as fast as you'd need: on my machine, it took about 30 seconds to search an array of 100 x 100 random numbers for $m = 3, $n = 4, and runtime will be O(n^2) the total number of integers in your list of sequences.

Also this finds only common subsequences of the exact length specified, so in this case it returns (5,10), (10,5), (6,21), (21,5). I can't see a way to adapt this to return directly only the longest common sequences, but you could maybe save all the length-2 sequences, then search for length-3 sequences and discard their subsequences, iteratively until you've found the longest subsequence.

Hugo

In reply to Re: Seeking algorithm for finding common continous sub-patterns by hv
in thread Seeking algorithm for finding common continous sub-patterns by johnnywang

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