Simple proof that GUID is not unique [closed]

Answer 1

Counting to 2^128 - ambitious.

Lets imagine that we can count 2^32 IDs per second per machine - not that ambitious, since it's not even 4.3 billion per second. Lets dedicate 2^32 machines to that task. Furthermore, lets get 2^32 civilisations to each dedicate the same resources to the task.

So far, we can count 2^96 IDs per second, meaning we will be counting for 2^32 seconds (a little over 136 years).

Now, all we need is to get 4,294,967,296 civilisations to each dedicate 4,294,967,296 machines, each machine capable of counting 4,294,967,296 IDs per second, purely to this task for the next 136 years or so - I suggest we get started on this essential task right now ;-)

Answer 2

Here's a nifty little extension method that you can use if you want to check guid uniqueness in many places in your code.

internal static class GuidExt
{
    public static bool IsUnique(this Guid guid)
    {
        while (guid != Guid.NewGuid())
        { }
        return false;
    }
}

To call it, simply call Guid.IsUnique whenever you generate a new guid...

Guid g = Guid.NewGuid();
if (!g.IsUnique())
{
    throw new GuidIsNotUniqueException();
}

...heck, I'd even recommend calling it twice to make sure it got it right in the first round.

Answer 3

Well if the running time of 83 billion years does not scare you, think that you will also need to store the generated GUIDs somewhere to check if you have a duplicate; storing 2^128 16-byte numbers would only require you to allocate 4951760157141521099596496896 terabytes of RAM upfront, so imagining you have a computer which could fit all that and that you somehow find a place to buy terabyte DIMMs at 10 grams each, combined they will weigh more than 8 Earth masses, so you can seriously shift it off the current orbit, before you even press "Run". Think twice!

Answer 4

If GUID collisions are a concern, I would recommend using the ScottGuID instead.

Answer 5

You can show that in O(1) time with a variant of the quantum bogosort algorithm.

Guid g1 = Guid.NewGuid();
Guid g2 = Guid.NewGuid();
if(g1 != g2) Universe.Current.Destroy();

Answer 6

The odds of a bug in the GUID generating code are much higher than the odds of the algorithm generating a collision. The odds of a bug in your code to test the GUIDs is even greater. Give up.