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Re: Analysing a (binary) string.

by ambidexter (Initiate)
on Jun 28, 2013 at 05:40 UTC ( #1041138=note: print w/ replies, xml ) Need Help??


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 !!


Comment on Re: Analysing a (binary) string.
Re^2: Analysing a (binary) string.
by hdb (Parson) 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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