We know that the "short" 32 bit OpenPGP key IDs can be easily brute-forced and the recommendation is to use the "long" 64 bit IDs or full 160 bit SHA1 fingerprint. However I am concerned that the "long" key IDs may also be brute-forceable in the not-so-distant future.

Given that a collision would generally be a precursor to a successful preimage attack, are there any known examples of a collision being found for a long key ID?


OpenPGP IDs are portions of the SHA-1 fingerprint, as defined in the standard. A short key ID is the last 32 bits, and the long key ID is the last 64 bits. A collision typically takes an average of 2n/2 operations, where n is the size of the hash in bits. Generating collisions for short IDs and long IDs is trivial, requiring an average of 216 and 232 operations, respectively. A collision in this case is defined as creating two differing hash inputs which have identical digests. It is actually even possible to collide the full SHA-1 fingerprint, as Google has shown. Colliding a full OpenPGP fingerprint requires nothing more than colliding a single SHA-1 hash. This is difficult, but possible with sufficient computing power.

Creating a preimage is different. Unlike a collision, a preimage attack requires creating an input that matches a specific hash digest. The attacker does not get to provide both inputs, only one, making it a much more difficult attack. Unlike a collision attack, a preimage attack requires a full 2n operations for a hash of size n. Because of this, creating a preimage for a short ID requires only 232 operations, which is downright trivial. Doing the same for a long ID requires 264 operations, which is not easy, but far from impossible. Doing the same with the full 160-bit fingerprint is simply impossible with current technology.

In summary:

|             | Collision |  Preimage  |
| Short ID    |  Trivial  |    Easy    |
| Long ID     |   Easy    |    Hard    |
| Fingerprint |   Hard    | Impossible |
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    Generic collision of 160-bit SHA-1 is 2^80, but the Google+Stevens et al break is 2^63.1. However it requires substantial freedom in fairly large inputs (they use PDF files), and I'm not sure if that can be obtained for (otherwise valid) PGP keys. – dave_thompson_085 Nov 23 '18 at 8:12
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    @dave_thompson_085 Specifically, IIRC, they had the freedom to manipulate the data in a non-displayed JPG image within the PDF. – TripeHound Nov 23 '18 at 9:53
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    @TripeHound Remember that the OpenPGP standard can hold a lot of non-displayed data that is free to be modified. – forest Nov 24 '18 at 1:25

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