it's not as difficult as you may think. the regex looks like this: /.*?([0-9]).*?\1/. including the brackets, it's 17 characters long! here's the full program:

#!/usr/local/bin/perl -w
use strict;
$|++;

my @numlist;
push @numlist, int(rand 100000) for 1..10;

for(@numlist) {
    next if /^$/;
    /.*?([0-9]).*?\1/ ? print "$_\tmultiple $1\n" : print "$_\tsingle\
+n";
}

[download]
i generated a list of random five digit numbers to feed the regex. here's a breakdown of the regex:
.*? i match zero or more characters, non-greedily
([0..9]) match numbers 0..9, save in \1
.*? again match zero or more characters, non-greedily
\1 find the next instance of the character in \1 *

*what's most interesting about this, is that $1 does not work in place of \1. instead, on my perl 5.6.1 install, it warns with: 

Use of uninitialized value in concatenation (.) or string at test_matc
+h1.pl line 10.
Nested quantifiers before HERE mark in regex m/.*?([0-9]).*?+ << HERE 
+/ at test_match1.pl line 10.

[download]
by the way, there's nothing to stop this from working with letters, as well. just change 0..9 and you're on your set.

enjoy!

Update: i see the error of my ways. tilly's right (below). i solved the exact opposite of the problem, which was pretty easy! whoops!

~Particle

In reply to Re: (GOLF) - multiple digit finder regex - 17 chars by particle
in thread regexp golf - homework by Boots111

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