in reply to Analysing a (binary) string. (Solved)

Hi, BrowserUk

If i understand it right, ! Since we dont know the length of the sub string that might repeat. It can be of length from 1-N/2. So if we start from 1 we need to parse the whole array N and for length 2 - (N-1) for length 3 - (N-2). So the time complexity of any algorithm for this problem would be "~N factorial". So if the string length is > 100 then it might take hours to get the complete solution !!

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Re^2: Analysing a (binary) string.
by hdb (Monsignor) on Jun 28, 2013 at 12:23 UTC

    I cannot see where you get N factorial from. To me it seems what you describe is N^2. This is also what I get from the brute force method below: 

    use strict;
use warnings;
use Time::HiRes qw/time/;

sub find_best_offset {
    my $strref = shift;
    my $n = length $$strref;
    my $mindiff = $n;
    my $best_offset = -1;
    for( my $i=1; $i<=$n/2; $i++ ) {
        # shift/rotate string and compare to original
        my $diff = $$strref ^ substr( $$strref, -$i ).substr( $$strref
+, 0, -$i );
        # number of differing characters between shifted string and or
+iginal
        my $ndiff = $diff =~ tr/\0//c; 
        # test for and keep optimal solution
        if( $ndiff < $mindiff ) {
            return $i unless $ndiff; # best case!
            $mindiff = $ndiff;
            $best_offset = $i;
        }
    }
    return $best_offset;
}

for ( 9, 99, 999, 9999, 99999 ) {
    my $t = time;
    my $str1 = "01234567890123456789".("01234567890123456x89" x $_);
    print find_best_offset( \$str1 );
    print "\t$_\t",time-$t,"\n";
}

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