Safety numbers in Signal are derived from a hash of the conversation's users public keys and their phone numbers. Safety number are used to ensure that the conversation was not MITM-ed.

When deriving safety numbers, SHA-512 iterated for 5200 times. According to the Signal safety blog, there were privacy issues re:phone numbers embedded in hashes. However this cannot be the reason, given the set of possible phone numbers is relatively small.

Comments in the source code:

The higher the iteration count, the higher the security level:

  • 1024 ~ 109.7 bits
  • 1400 > 110 bits
  • 5200 > 112 bits

So: what is the reason for intentionally slowing down the Safety Numbers computation?

Bonus: how are roughly the security levels (1024 SHA-512 hashes ~ 109.7 bits) computed?

  • 5
    +1 Given that NIST recommends 10,000 iterations of SHA2 for PBKDF2, I think you're right to challenge 5200 > 112 bits. – Mike Ounsworth Oct 15 '18 at 16:11

The comment isn't explained very well, but I believe I've determined the math behind it. The safety number is 60 base 10 digits, but it's created in two 30 digit halves: one based on your phone number and public key, and one based on the phone number and public key of the person you're talking to.

Assuming a high entropy value is converted into a 30 digit number without unnecessary loss of entropy, it will contain log2(1030) ≈ 99.66 bits of entropy, equating to 99.66 "bits of security" (meaning an attacker would have a 50% chance of matching that safety number after 299.66/2 = 298.66 hashes). Iterating many times increases the bits of security (since it increases the number of hash operations for each try the attacker makes):

log2(1030 × 1024) ≈ 109.66

log2(1030 × 1400) ≈ 110.11

log2(1030 × 5200) ≈ 112.00

This is for how many hashes an attacker would have to perform to match a specific security number, but if the attacker wanted to be able to read the messages you send them, they'd need to know the private key that matches the public key they used in the hash. Generating RSA keypairs is computationally expensive, but ECC is much faster. If the keypair generation is fast enough, it makes sense to iterate the hash to increase the lower bound of an attack on a safety number.

  • Makes you wonder why they didn't just make the signature 40 digit. – James_pic Oct 16 '18 at 16:34

The safety number is a derivation of the stable identifier and the public key of a user. Safety numbers are computed for both people in a conversation.

The real important code is this snipit

byte[] publicKey = getLogicalKeyBytes(unsortedIdentityKeys);  
byte[] hash = ByteUtil.combine(ByteUtil.shortToByteArray(FINGERPRINT_VERSION), publicKey, stableIdentifier.getBytes());  

for (int i=0;i<iterations;i++) {
        hash = digest.digest(publicKey);

What's happening in is we are taking the fingerprint version, public key, and stable identifier as starting inputs and hashing that once with SHA-512. The second iteration apends the public key to the hash we just produced, then hashes it a second time.

This process of adding the public key and repeating the hash continues for the number of indicated iterations.

Why do we need to do more iterations than the past?

This is due to a fundamental issue if hashing. Which is the possibility of hash collisions.

Let's say I'm an attacker (Eve). Alice wants to talk to Bob, so Signal sends her public key to Bob, but Eve intercepts the public key and substitutes her own. Normally there is an indication the key changed, and the Safety Number changes.

IF Eve had enough resources she could construct a public key which matched the safety number. To combat this threat we make it so that Eve would need to find a collision which occurs after 5200 rounds of hashing, with adding that same key every round.

This becomes computationally infeasible since each round of hashing makes finding a collision linearly more computationally expensive. The number of iterations currently picked usually is calculated on how long an attack of this style would take based in resources of the percieved threat.

I can't find any calculations from Signal as to specifically why they picked 5200.

  • 1
    log2(10^30 * 5192) = 111.999918 and log2(10^30 * 5193) = 112,0001954. 5200 was probably taken to have a "smooth" number to calculate 112 bits of entropy. – Tom K. Oct 16 '18 at 11:30
  • hello sir, anywhere i can contact you @Daisetsu – turmuka Dec 15 '18 at 8:09
  • @turmuka do you have a comment about this particular answer? – Daisetsu Dec 16 '18 at 10:45

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