4 ms·

I'm not sold on this argument. Why is a 40-char long fingerprint better that verifying the last 40 chars of the public key?

• otachack 3y ago
  Because of tampering. If an attacker can produce a pair where the public key's last 40 chars match the victim's public key last 40 chars they effectively have a public key to dish out via MITM. How feasible it is to produce said pair is another story.
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• tux3 3y ago
  It doesn't sound too hard to generate an RSA "vanity key", with any value you want for some of the bytes. You can't control _all_ of the bytes, because it still needs to have the right structure and for you to have the corresponding private key, but 40 bytes of your choosing seems completely doable. And if you can do that, you can impersonate someone else whose pubkey has the same 40 bytes. With a hash, any bit difference in any part of the key should result in a completely different fingerprint (hash collisions being extremely hard to find).
  • lucb1e 3y ago
    Note that this is specific to RSA keys. In no scenario (that doesn't involve extraterrestrial resources or perhaps nuclear fusion) can you create an ECC key, let alone a hash fingerprint, with 40 bytes of vanity. Afaik ECC keys are considered to be half the strength of a symmetric key (pre-quantum), so that's 20 fully random bytes or 160 bits of entropy. The sun simply doesn't hit the earth with enough energy even if you'd capture 100% of that and starve all life for it to do a computation of that magnitude. (The boundary is around the standard 128-bit key size iirc; I keep forgetting if the sun's energy would be sufficient for like ~120 bits or more like ~140 bits... and that's assuming perfectly efficient computational machines.) Since the person you're responding to didn't specify which public key type, and since it's not obvious that your mention of RSA is to the exclusion of other algorithms such as ECC, I felt like the comment is a bit misleading
• eadmund 3y ago
  Because there is less entropy in 40 characters of public key than there is in 40 characters of cryptographic hash of a public key.
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  • brookst 3y ago
    Great answer, but why? Doesn’t that suggest the public key should be suitable for lossless compression to fewer bits?
    • smnc 3y ago
      Hashing is not compression, it should not be reversible/inflatable.
    • Dylan16807 3y ago
      You have to leak information about the key generation to compress it.
    • lucb1e 3y ago
      I think the two existing replies misunderstood your question, but it's a good one. Which is to say, I don't know the answer but I feel like I should! I'll take a guess. There are large gaps between good RSA keys. 100 may be a valid key and 138, but not anything in between. Or, well, they're valid but they're trivially broken by having a divisor other than 1, itself, and the huge prime factors (the example of 100 and 138 are not good keys for exactly that reason; finding secure keys is left as an...). That's why we need RSA keys that are more than 256 bits in length: the key space is sparse and an attacker can, with some amount of efficiency, skip over the gaps. (This is all from years-old memories of how RSA works, don't take this for absolute certainty.) What I'm guessing the answer to your question is, is this: it must be inefficient to reconstruct the key from an indexed form (e.g.: the first good key (100) has index 1, the second good key (138) index 2, etc.) without spending computational power disproportionate to the amount of extra resources that storing/transmitting the full key takes. Now that I read the other answers again, maybe that's what Dylan meant, but to me that answer seems wrong because the public key is argued to not be uniformly random and that's precisely what compression algorithms are able/made to deal with. Perhaps not as efficiently as indexing can, but still. You wouldn't need to apply it to the prime factors or private key (doing that would, as they say, leak information), just the public part which people were saying is not fully random.
• petra 3y ago
  One possible public key is zeroes + public fingerprint. If I remember correctly diffie-helman is based on multiplications , so maybe finding the private key is now equivalent to finding the private key of a 40 chat public key, which may be doable. I'm probably wrong on the details here , but there's probably some math tricks you could use to more easily find some private, public key pairs that end with the fingerprint.
• wiml 3y ago
  It's really easy to generate a public RSA key with desired patterns in it. The 1992-era PGP did use the last few bytes of the public key as the identifier, but later versions moved to using a truncated hash (MD5, and later SHA). At some point someone generated colliding keys for all the keys in the public keyring and uploaded them all, which kinda drove the point home. (I assume that it's harder to generate a public ECDSA key with a specific pattern, but elliptic curve stuff didn't become common until after hashes were used for key identifiers.)
• rainsford 3y ago
  Others have answered this question pretty well for large public keys, but I wonder about the same argument for shorter elliptic curve keys where the key length and hashed fingerprint length may be the same. For example, is comparing 8 bytes of a Curve25519 public key as good as comparing 8 bytes of a SHA256 hash of that key? My gut says no, since generating partial cryptographic hash collisions is completely random while a structured public key of any kind is presumably less random, but I'm not sure how much less random (and thus easier) it would be.
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