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Comment by csense | Hacker News Reader

What's preventing you from just picking at random an ID that hasn't been assigned?

If you're a small-volume project, then presumably people will need to have custom kernel patches anyway and will be unlikely to actually have a device with a conflicting ID.

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Nothing, which is why it is arguably a short-sighted move for USB-IF to charge such a large amount to hobbyists and such.

I'm surprised that the USB-IF itself doesn't have several hobbiest vendor ids that then hand out single PIDs for use (say at $25USD)

This would lead to potential ID collisions as the VID space isn't very big. So just use a well-known debugging ID and have your driver detect your device using other USB descriptor fields.

Given that the full identification of a device is a VID:PID pair, and that's a 32-bit space, picking one at random is not very likely to result in a collision. Furthermore, this assumes true randomness and not "randomly pick one and then check online to see if some other device has already claimed it."

It also assumes that your project isn't close enough to some existing one that you really need to define a completely new interface instead of just using or extending that of an existing device.

People tend to confuse the odds of two arbitrary strings colliding with the chances of a set of strings having a collision. 32bit space is not that big, and you have to take into account the birthday problem.

If you do out the math it turns out there's a 50% chance of collision after only 77,163 hashes.

http://preshing.com/20110504/hash-collision-probabilities/

A good rough rule is to halve the power of two: so 2^16 = 65536, which is close enough to 77163.

That's really cool. I tried it with the birthday problem[1] (real answer 23) and got 16 if I used whole exponents[2] and 19 with fractional exponents[3]. They're a little off but it gives you an idea of the value.

[1] https://en.wikipedia.org/wiki/Birthday_problem

[2] 256 days=2^8, 16=2^4

[3] 365 days=2^8.51, 19=2^4.255

That doesn't matter to an individual developer. Why should they care if the set of all in use USB PIDs has collisions? It's only becomes a problem for them if their PID suffers a collision.

The Birthday Problem is only relevant to find collisions between any 2 strings, but in this case the developer is only concerned about the probability of his own string clashing with someone else’s, thus the probability of a collision is only Number of taken strings / Number of total strings.

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