Matching a pattern across a 32767-length boundary is probably not the best solution. It feels like looking looking for nails because you have a hammer. Here's a solution that avoids quantifiers. I'm not going to take the time to test the speed of this algorithm. If it's not fast enough, drop into Inline::C and prefer a non-regex means of detecting boundaries.

use strict;
use warnings;

for (1 .. 25) {
    my $digits = join '', map {int rand 2} 1 .. 100_000;
    my ($pos, $len) = longest($digits);
    my $substr = substr($digits, $pos, $len);
    printf "%02d: At position (%6d), length (%6d) => $substr\n", $_, $
+pos, $len;
}

sub longest {
    my ($seq, $lastpos, $maxpos, $maxlen) = (shift, 0, 0, 0);

    while ($seq =~ m/(.)(?!\1)/g) {

        my $p    = pos $seq;
        my $len  = $p - $lastpos;
        $lastpos = $p;

        if ($len > $maxlen) {
            $maxlen = $len;
            $maxpos = $p-$len;
        }
    }

    return $maxpos, $maxlen;
}

__END__

01: At position ( 54044), length (    18) => 111111111111111111
02: At position ( 22821), length (    15) => 000000000000000
03: At position ( 84563), length (    18) => 111111111111111111
04: At position ( 97707), length (    18) => 000000000000000000
05: At position (  2567), length (    16) => 1111111111111111
06: At position ( 31038), length (    18) => 111111111111111111
07: At position ( 73339), length (    16) => 0000000000000000
08: At position ( 26644), length (    19) => 0000000000000000000
09: At position (  9906), length (    21) => 111111111111111111111
10: At position ( 50662), length (    15) => 111111111111111
11: At position ( 86843), length (    15) => 111111111111111
12: At position ( 15995), length (    16) => 0000000000000000
13: At position (  4399), length (    15) => 000000000000000
14: At position ( 25401), length (    19) => 0000000000000000000
15: At position ( 65784), length (    21) => 000000000000000000000
16: At position ( 37043), length (    14) => 00000000000000
17: At position ( 63608), length (    18) => 111111111111111111
18: At position ( 69870), length (    17) => 00000000000000000
19: At position ( 20108), length (    18) => 000000000000000000
20: At position ( 33099), length (    19) => 1111111111111111111
21: At position ( 40355), length (    17) => 00000000000000000
22: At position ( 98429), length (    18) => 000000000000000000
23: At position ( 43568), length (    16) => 1111111111111111
24: At position ( 71050), length (    16) => 0000000000000000
25: At position ( 45697), length (    19) => 0000000000000000000

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This scales up to any string size and any greatest-common-substring size. In other words, '1' x (100000) is not a problem. By doing away with the use of the regex altogether, and by eliminating the use of substr, it could even be adapted to work with infinite streams of digits.


Dave

In reply to Re^5: How to match more than 32766 times in regex? by davido
in thread How to match more than 32766 times in regex? by rsFalse

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