Home/ Questions/Q 8535411
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-06-11T10:25:28+00:00

I’m implementing some program which uses id’s with variable length. These id’s identify a

  • 0

I’m implementing some program which uses id’s with variable length. These id’s identify a message and are sent to a broker which will perform some operation (not relevant to the question). However, the maximum length for this id in the broker is 24 bytes. I was thinking about hashing the id (prior to sending to the broker) with SHA and removing some bytes until it gets 24 bytes only.

However, I want to have an idea of how much will this increase the collisions. So this is what I got until now:

I found out that for a “perfect” hash we have the formula p^2 / 2^n+1 to describe the probability of collisions and where p is the number of messages and n is the size of the message in bits. Here is where my problem starts. I’m assuming that removing some bytes from the final hash the function still remains “perfect” and I can still use the same formula. So assuming this I get:

 5160^2 / 2^192 + 1 = 2.12x10^-51

Where 5160 is the pick number of messages and 192 is basically the number of bits in 24 bytes.

My questions:

  • Is my assumption correct? Does the hash stay “perfect” by removing some bytes.

  • If so and since the probability is really small, which bytes should I remove? Most or less significant? Does it really matter at all?

PS: Any other suggestion to achieve the same result is welcomed. Thanks.

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    However, the maximum length for this id in the broker is 24 bytes. I was thinking about hashing the id (prior to sending to the broker) with SHA and removing some bytes until it gets 24 bytes only.

    SHA-1 outputs only 20 bytes (160 bits), so you’d need to pad it. At least if all bytes are valid, and you’re not restricted to hex or Base64. I recommend using truncated SHA-2 instead.

    Is my assumption correct? Does the hash stay “perfect” by removing some bytes.

    Pretty much. Truncating hashes should conserve all their important properties, obviously at the reduced security level corresponding to the smaller output size.

    If so and since the probability is really small, which bytes should I remove? Most or less significant? Does it really matter at all?

    That should not matter at all. NIST defined a truncated SHA-2 variant, called SHA-224, which takes the first 28 bytes of SHA-256 using a different initial state for the hash calculation.

    My recommendation is to use SHA-256, keeping the first 24 bytes. This requires around 2^96 hash-function calls to find one collision. Which is currently infeasible, even for extremely powerful attackers, and essentially impossible for accidental collisions.

    • 0