[Cryptography] SHA-256 decrypted (8 rounds)

[Ray Dillinger]

On 11/8/23 03:25, McDair wrote:
>
> As stated, there is no guessing involved here, meaning a fixed number 
> of iterations that will lead to the/a valid input message.
>
This wouldn't be true if your code is generating a preimage 
(reconstructing the input).  In what sense, specifically, are you using 
the word 'decrypt'?  Because in precise usage it does not have a defined 
meaning with respect to a hash function, so it's not clear what you're 
claiming.

Are you assuming that the input which you're trying to reconstruct is 
also a single block, or at any rate shorter than the hash?  That's 
what's required for there to be a unique solution which is what you're 
claiming when you say 'no guessing' - but in that case it wouldn't be a 
hash function at all.

> Because the SHA-256 output hash still fits in a single input block, 
> the same 'decryption'/reversion method (limited to 8 rounds here) can 
> also be used for SHA-256D (by also applying it twice). Or a multitude 
> of hashes of hashes for that matter.
>
Again the notion of applying a hash to a single block message. That 
doesn't make any sense.

For that matter I don't think your VB code works for SHA-256 at all if 
implementation of the hash rounds are the same as the published 
algorithm.  It looks like you are generating a hash of a single-block 
input rather than generating a preimage (of any length) given a 
single-block output.

Bear

-------------- next part --------------
An HTML attachment was scrubbed...
URL: <https://www.metzdowd.com/pipermail/cryptography/attachments/20231109/6523139d/attachment.htm>