Ok you are right,

so because i have no patience at all with regexes you can exploit the fact that your valid strings are always odd; they start and end with x and another x must be in the middle.

use strict; use warnings; while (<DATA>){ chomp; # note the string IS always odd my $inter = int ((length $_) / 2)-1; my @char = $_=~/./g; if (scalar @char % 2 < 1){ print "Not OK $_ (unbalanced)\n"; next; } if ( $char[0] eq $char[$inter+1] and $char[0] eq $char[-1] and $char[0] eq 'x' ){ print "$_\t\tOK\n"; } else { print "NOT OK $_\t[$char[0] $char[$inter+1] $char[-1]]\n"; } } __DATA__ xxxxx x1x2x...x xxx x.x.x x12x..x x123x...x x123x.x.x x12x1x # out xxxxx OK x1x2x...x OK xxx OK x.x.x OK x12x..x OK x123x...x OK x123x.x.x OK Not OK x12x1x (unbalanced)

L*

UPDATE: it can be semplified, or golfed, a lot using 5.010

use strict; use warnings; use 5.010; while (<DATA>){ chomp; # note the string IS always odd if ((length $_) % 2 < 1){ print "Not OK $_ (unbalanced)\n"; next; } if (($_=~/./g)[0,(int((length $_)/2)-1),-1]~~[qw(x x x)]){ print "$_\t\tOK\n"; } else { print "NOT OK $_\n"; } }

L*

There are no rules, there are no thumbs..
Reinvent the wheel, then learn The Wheel; may be one day you reinvent one of THE WHEELS.

In reply to Re^3: Regular Expression: search two times for the same number of any signs -- no regex at all (updated) by Discipulus
in thread Regular Expression: search two times for the same number of any signs by Anonymous Monk

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