SHA-256 by hand

Question

Yes, it would take days. I am not in a rush, nor trying to solve any problem...I just like Maths.

There is a video floating around (citation/source to come) of someone doing the first round of SHA-256 by hand. It seems straight forward, but I have no idea what the second round looks like. Do I use the old values from the last round? Do I use more chunks of the plaintext?

Please don't just quote Wikipedia pseudocode at Me...I already read it. Did. Not. Help.

I'll update when I find a link to the video. Thanks.

Answer 1

Yes, the next round use the old values from the last round.

For the first 16 out of 64 rounds, the next round does use the next 32-bit chunk of plaintext (message to hash) when that is 512-bit or more; but things are more complex in general, because what's processed in a round is a block of the padded message rather than the actual message, and that block undergoes transformations that matter for the next 48 out of 64 rounds.

SHA-256 can be described as follows:

• setup a vector of 8 32-bit registers $S[\,]$ to an certain initial value
• for each of $\lceil(L+65)/512\rceil$ 512-bit block(s) of the padded message to hash (consisting of the message of length $L$ bit(s), a single 1 bit, $511-(L+65\bmod 512)$ 0 bit(s), and the 64-bit representation in big-endian binary of $L$ )
  • copy that padded block of plaintext into a vector of 16 32-bit registers $W[\,]$ per big-endian binary
  • copy the whole of $S[\,]$ into $T[\,]$
  • for each of 64 rounds $r$ from $0$ to $63$
    • if $r\ge16$
      • transform $W[\,]$ in a certain way involving 4 rotations, 2 shifts, 4 XOR, and 3 additions modulo $2^{32}$ (all over 32-bit quantities)
    • transform $S[\,]$ in a certain way involving the 32-bit value $W[r\bmod16]$ (thus coinciding with padded message for the first 16 out of 64 rounds), a 32-bit constant $K[r]$, 6 rotations, 4 XOR, 7 additions modulo $2^{32}$, and the evaluation of the choice and majority ternary bitwise functions (all over 32-bit quantities)
  • add each of the 8 components of $T[\,]$ into $S[\,]$, modulo $2^{32}$
• the hash is $S[\,]$ with each 32-bit register expressed per big-endian binary, forming a 256-bit bitstring.