Date: Fri, 11 Aug 2006 03:42:03 -0700 From: "Nikolas Britton" <nikolas.britton@gmail.com> To: "Marc G. Fournier" <scrappy@freebsd.org> Cc: Paul Schmehl <pauls@utdallas.edu>, freebsd-questions@freebsd.org Subject: Re: BSDstats Project v2.0 ... Message-ID: <ef10de9a0608110342q62f81fc8p5fb4b4df37595593@mail.gmail.com> In-Reply-To: <ef10de9a0608100327r5b402d64xc4eef38a4f61ba4e@mail.gmail.com> References: <20060807003815.C7522@ganymede.hub.org> <20060808201359.S7522@ganymede.hub.org> <44D91F02.90107@mawer.org> <20060808212719.L7522@ganymede.hub.org> <20060809072313.GA19441@sysadm.stc> <20060809055245.J7522@ganymede.hub.org> <44D9F9C4.4050406@utdallas.edu> <20060809130354.U7522@ganymede.hub.org> <ef10de9a0608091700x6cc268ear6566c26f93f1fdf0@mail.gmail.com> <ef10de9a0608100327r5b402d64xc4eef38a4f61ba4e@mail.gmail.com>
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Ok... With my new script it took only 158 minutes to compute ALL
TCP/IP address hashes. I'll repeat that... I have an md5 hash for
every IP address in the world! All I need to do is grep your hash and
it will tell me your IP address. yippee! :-)
Can we please find a new method to track hosts... perhaps my earlier
example: ifconfig |md5. If not please remove my entries in the
database.
I've attached the script used to make the hashes.
On 8/10/06, Nikolas Britton <nikolas.britton@gmail.com> wrote:
> On 8/9/06, Nikolas Britton <nikolas.britton@gmail.com> wrote:
> > On 8/9/06, Marc G. Fournier <scrappy@freebsd.org> wrote:
> > > On Wed, 9 Aug 2006, Paul Schmehl wrote:
> > >
> > > > Marc G. Fournier wrote:
> > > >> On Wed, 9 Aug 2006, Igor Robul wrote:
> > > >>
> > > >>> On Tue, Aug 08, 2006 at 09:30:42PM -0300, Marc G. Fournier wrote:
> > > >>>> Could create problems long term .. one thing I will be using the
> > > >>>> IPs to do is:
> > > >>>>
> > > >>>> SELECT ip, count(1) FROM systems GROUP BY ip ORDER BY count DESC;
> > > >>>>
> > > >>>> to look for any 'abnormalities' like todays with Armenia ...
> > > >>>>
> > > >>>> hashing it would make stuff like that fairly difficult ...
> > > >>> You can make _two_ hashes and then concatenate to form unique key.
> > > >>> Then you still be able to see "a lot of single IPs". Personaly, I dont
> > > >>> care very much about IP/hostname disclosure :-)
> > > >>
> > > >> Except that you are disclosing that each and every time you send out an
> > > >> email, or hit a web site ... :)
> > > >>
> > > > The systems I'm concerned about are on private IP space, to not send email
> > > > and don't have X installed, much less a web browser and can only access
> > > > certain FreeBSD sites to update ports. In fact, they're not even accessible
> > > > from *inside* our network except from certain hosts. In order to
> > > > successfully run the stats script on these hosts, I would have to open a hole
> > > > in the firewall to bsdstats.hub.org on the correct port.
> > > >
> > > > And yes, I *am* paranoid. But if you really want *all* statistics you can
> > > > get, then you'll have to deal with us paranoid types. My workstation, which
> > > > is on a public IP, is already registered.
> > >
> > > Done ... now I really hope that the US stats rise, maybe? I have a hard
> > > time believing that Russia and the Ukraine have more deployments then the
> > > 'good ol'US of A' ... or do they? *raised eyebrow*
> > >
> > > Here is what is now stored in the database (using my IP as a basis)
> > >
> > > # select * from systems where ip = md5('24.224.179.167');
> > > id | ip | hostname | operating_system | release | architecture | country | report_date
> > > ------+----------------------------------+----------------------------------+------------------+------------+--------------+---------+---------------------------
> > > 1295 | 45c80b9266a5a6683eee9c9798bd6575 | 4a9110019f2ca076407ed838bf190017 | FreeBSD | 6.1-RC1 | i386 | CA | 2006-08-09 02:34:05.12579
> > > 1 | 45c80b9266a5a6683eee9c9798bd6575 | 9a45e58ab9535d89f0a7d2092b816364 | FreeBSD | 6.1-STABLE | i386 | CA | 2006-08-09 16:01:03.34788
> > >
> >
> > Why don't you just broadcast the ip address, it's what your doing now
> > anyways. 253^4 is a very small number.
> >
> > infomatic# perl
> > my $num = 0;
> > system "date";
> > while ($num <= 409715208) {
> > $num++
> > }
> > system "date";
> > Wed Aug 9 18:18:45 CDT 2006
> > Wed Aug 9 18:20:48 CDT 2006
> >
> > 2 minutes * 10 = 20 minutes to iterate though 4 billion IP addresses
> > on a very slow uni-proc system. I could even store every IP to md5
> > hash using less then 222GB of uncompressed space.
> >
> > If you want... give me the md5 hash of a real ip address that is
> > unknown to me and I will hand you the ip address in two days... or
> > less. run the IP address though like this:
> >
> > md5 -s "xxx.xxx.xxx.xxx"
> >
> > I have other things to do with my time, so I don't really want to do
> > this, but if that's what it takes to stop this idea dead I'll do it.
> >
> >
>
> Here's a better way to explain the problem:
>
> Let's say we need to find Marc's IP address but we only have it's md5
> hash value. Some of you may think this is hard to do but it's not. All
> we need to do is compute every IP address into a hash and then match
> Marc's hash to one in are list:
>
> 24.224.179.164 = e7e7a967c5f88d9fb10a1f22cd2133d2
> 24.224.179.165 = 3aa9b50aa7190f5aca1f78f075dc69c2
> 24.224.179.166 = c695175e48d649e3496ac715406a488d
> 24.224.179.167 = 45c80b9266a5a6683eee9c9798bd6575
>
> So what is an IP address?... mathematically speaking it's 4 base 255
> numbers grouped together:
>
> {0, ..., 255}.{0, ..., 255}.{0, ..., 255}.{0, ..., 255}
>
> To calculate how many combinations there could be you simply take the
> base unit and raise it to the 4th power, since there are 4 of them.
> This gives us 255^4 combinations or 4,228,250,625 TCP/IP addresses. We
> also know that the first number can't be 0 or 255 and the others can't
> be 255, we can also rule out all 127.x.y.z loopback and multicast
> 224.x.y.z - 239.x.y.z addresses:
>
> (237^1) * (254^3)
>
> This leaves us with 3,883,734,168 valid IP addresses. We can divide
> this number by 5,000 and run it through a simple perl script to get a
> time estimate on how long it will take to compute all these hashs. We
> will split it into 4 parallel jobs:
>
> my $number = 0;
> while ($number <= 194187) {
> system "md5 -s $number >> /usr/data/hashlist1";
> $number++;
> }
>
> my $number = 194188;
> while ($number <= 388373) {
> system "md5 -s $number >> /usr/data/hashlist2";
> $number++;
> }
>
> my $number = 388374;
> while ($number <= 582560) {
> system "md5 -s $number >> /usr/data/hashlist3";
> $number++;
> }
>
> my $number = 582561;
> while ($number <= 776747) {
> system "md5 -s $number >> /usr/data/hashlist4";
> $number++;
> }
>
> Ok, it took me 48 minutes to go though 1/5000th of the numbers using 1
> dual-core Xeon system. Considering that my algorithm is very
> inefficient and that this task is perfect for cluster computing we can
> easily beat this time estimate...
>
> If a cracker got hold of your database he could crack all the hashes
> in a short amount of time and then have the IP addresses, detailed
> system version, and full hardware information for exploitation. very
> very bad. For hashes to work correctly the input data needs to be
> larger then the hash itself, for example:
>
> asglkhasdlgkjhasldkjhadlkjfhadlgkjhsadlkgjhsadlaskjdhgqalsdjkh
> in md5 is e498d452efdfbfda87e522ff3af3b638. To crack that you have to
> tackel ether the hash itself or the hash input... both are extremely
> large numbers and impossible to brute force using todays hardware:
>
> 16^32 is the md5 hash, hexadecimal is base 16:
> 16^32 = 340,282,366,920,938,463,463,374,607,431,768,211,456
>
> 26^62 is the input for the hash, {a, ..., z} is 26 letters:
> 26^62 = 53,499,235,273,362,382,919,605,484,267,229,0
> 01,477,944,554,670,188,588,661,044,895,155,128,259,48,95
> 0,330,580,402,176
>
--
BSD Podcasts @:
http://bsdtalk.blogspot.com/
http://freebsdforall.blogspot.com/
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