Simple proof that GUID is not unique [closed]

Answer 1

A GUID is theoretically non-unique. Here's your proof:

• GUID is a 128 bit number
• You cannot generate 2^128 + 1 or more GUIDs without re-using old GUIDs

However, if the entire power output of the sun was directed at performing this task, it would go cold long before it finished.

GUIDs can be generated using a number of different tactics, some of which take special measures to guarantee that a given machine will not generate the same GUID twice. Finding collisions in a particular algorithm would show that your particular method for generating GUIDs is bad, but would not prove anything about GUIDs in general.

Answer 2

Aren't you all missing a major point?

I thought GUIDs were generated using two things which make the chances of them being Globally unique quite high. One is they are seeded with the MAC address of the machine that you are on and two they use the time that they were generated plus a random number.

So unless you run it on the actual machine and run all you guesses within the smallest amount of time that the machine uses to represent a time in the GUID you will never generate the same number no matter how many guesses you take using the system call.

I guess if you know the actual way a GUID is made would actually shorten the time to guess quite substantially.

Tony

Answer 3

Not to p**s on the bonfire here, but it does actually happen, and yes, I understand the joking you have been giving this guy, but the GUID is unique only in principle, I bumped into this thread because there is a bug in the WP7 emulator which means every time it boots it gives out the SAME GUID the first time it is called! So, where in theory you cannot have a conflict, if there is a problem generating said GUI, then you can get duplicates

http://forums.create.msdn.com/forums/p/92086/597310.aspx#597310

Answer 4

Of course GUIDs can collide. Since GUIDs are 128-bits, just generate 2^128 + 1 of them and by the pigeonhole principle there must be a collision.

But when we say that a GUID is a unique, what we really mean is that the key space is so large that it is practically impossible to accidentally generate the same GUID twice (assuming that we are generating GUIDs randomly).

If you generate a sequence of n GUIDs randomly, then the probability of at least one collision is approximately p(n) = 1 - exp(-n^2 / 2 * 2^128) (this is the birthday problem with the number of possible birthdays being 2^128).

   n     p(n)
2^30 1.69e-21
2^40 1.77e-15
2^50 1.86e-10
2^60 1.95e-03

To make these numbers concrete, 2^60 = 1.15e+18. So, if you generate one billion GUIDs per second, it will take you 36 years to generate 2^60 random GUIDs and even then the probability that you have a collision is still 1.95e-03. You're more likely to be murdered at some point in your life (4.76e-03) than you are to find a collision over the next 36 years. Good luck.

Answer 5

1. Go to the cryogenics lab in the New York City.
2. Freeze yourself for (roughly) 1990 years.
3. Get a job at Planet Express.
4. Buy a brand-new CPU. Build a computer, run the program, and place it in the safe place with an pseudo-perpetual motion machine like the doomsday machine.
5. Wait until the time machine is invented.
6. Jump to the future using the time machine. If you bought 1YHz 128bit CPU, go to 3,938,453,320 days 20 hours 15 minutes 38 seconds 463 ms 463 μs 374 ns 607 ps after when you started to run the program.
7. ...?
8. PROFIT!!!

... It takes at least 10,783,127 years even if you had 1YHz CPU which is 1,000,000,000,000,000 (or 1,125,899,906,842,624 if you prefer to use binary prefix) times faster than 1GHz CPU.

So rather than waiting for the compute finished, it would be better to feed pigeons which lost their home because other n pigeons took their home. :(

Or, you can wait until 128-bit quantum computer is invented. Then you may prove that GUID is not unique, by using your program in reasonable time(maybe).

Answer 6

Here's a solution, too:

int main()
{
  QUuid uuid;
  while ( (uuid = QUuid::createUuid()) != QUuid::createUuid() ) { }
  std::cout << "Aha! I've found one! " << qPrintable( uuid.toString() ) << std::endl;
}

Note: requires Qt, but I guarantee that if you let it run long enough, it might find one.

(Note note: actually, now that I'm looking at it, there may be something about the generation algorithm that prevents two subsequently generated uuids that collide--but I kinda doubt it).