Score:0

AES in binary-circuit-based 2PC

fr flag

No matter AES with CBC or EBC etc, the inputs for AES i.e, plaintext, key, IV, are always provided by one party. In other words, there is no need for AES in a 2PC scenario since one party can already calculate and get the result.

When I check e.g. the AES-no-expanded.txt files, party one has 128 bits input and party two has also 128 bits input. The output is 128 bits. I assume the first 128 bits are for plaintext, the other 128 bits stand for IV or keys.

My questions are following:

  1. Why does AES exist for 2PC since one party is enough to calculate the result?
  2. Which type of AES does on their pages stand for? Since AES with CBC needs three inputs.
Score:2
us flag

Why does AES exist for 2PC since one party is enough to calculate the result?

The whole point of 2PC is that the two parties have different pieces of the input. If I have $K$ and you have $M$, and we want to keep both secret, then no single person has enough information to compute $\textsf{AES}(K,M)$. We need 2PC to do it.

Which type of AES does on their pages stand for? Since AES with CBC needs three inputs.

AES is a block cipher. It takes a key and data block as input. CBC is an encryption scheme, not a block cipher. CBC is a higher-level construction, taking other inputs (depending on exactly how you define things), that uses a block cipher like AES as a component. These circuits are for plain AES.

Willi avatar
fr flag
I assume K is part of plaintext and M is also part of plaintext. Who will provide the key in 2PC? If one party needs to provide key and plaintext, will it have 256 bits inputs?
us flag
$K$ = key. $M$ = plaintext
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