# Tag Info

12

Splitting a key does not reduce the key strength at all. Simply generate two random 128-bit strings and give one to each party. Encrypt the data with the exclusive OR of the two random strings. Each string alone gives no information whatsoever about the final key, assuming your random number generator is sound. No party has any advantage.

9

Mathematically, it can probably be done. There has been research into trapdoor block ciphers. See, e.g., A family of trapdoor ciphers by Rijmen and Preneel, and follow-up papers. In practice, though, the problem statement is not realistic. The assumptions are just not realistic. Today, there's no reason why Red would be limited to using Blue's ciphers. ...

6

The design of DES might give some insight into the problem. The NSA altered the S-box of DES. Many people thought they planted a backdoor. It wasn't until later that differential cryptanalysis was independently discovered by Biham and Shamir that people realized that the NSA actually made DES stronger. So the lesson to learn from this is: clearly the NSA ...

5

The two most popular ways I am aware of are Shamir secret sharing and additive secret sharing. I'll explain both. Additive Secret Sharing I'll start with additive as it is conceptually simpler (but also more limited). I'll also use bitwise addition modulo 2 as the addition operation (i.e., XOR), but know that that isn't the only option. You could use real, ...

4

The usual method of "splitting a key" $K$ into two keys $K_0$ and $K_1$ given to two parties, for the purpose described, is as follows: generate $K_1$ randomly, of the same size as $K$; set $K_0 = K \oplus K_1$. "Joining" $K_0$ and $K_1$ into $K$ is simply $K = K_0 \oplus K_1$. This construct is such that each of the two parties gain absolutely no ...

3

One simple way would be to use Trivial secret sharing: Split DOCUMENT in half into DOCUMENT1 and DOCUMENT2. Choose BITS1 and BITS2 uniformly and independently at random from binary strings of length equal to the lengths of DOCUMENT1 and DOCUMENT2, respectively. Give BITS2 and (BITS1 xor DOCUMENT1) to not-quite-trusted 3rd party A, and BITS1 and (BITS2 xor ...

3

You seem to be assuming that you will take the key, and give half the key bits to each party. While this can be made to work, there are cleverer ways to do this: You can pick an N bit random number (where N is the length of the key), and give one side that random number, and the other that random number xor'ed with the key. That way, neither side has any ...

2

In the asymmetric encryption context I think something can be done in this direction with a double trapdoor function. I studied few examples of them in the past and briefly you can build up an encryption scheme with a "local" trapdoor and a "global" one. If you keep the secret for the global trapdoor for you you'll have a sort of escrow key allowing you to ...

1

Can Alice obtain the session key due to the multiplicative properties of the modulus function and the basis of which RSA is built on? Essentially, yes. One way of looking why RSA works (that is, why the encryption and decryption are inverses of each other) is because of two mathematical identities: $$(M^a \bmod N)^b \bmod N = M^{a \cdot b} \bmod N$$ \...

1

It may looks like off-topic. But there exist hardware solution today. In the early years, both Soviet and USA used reverse engineering to get the scheme of crypto-system. Military standards today are too closed. What we use called a civil standard, and it is open. So reverse engineering seems unnecessary. There is such technology: ...

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