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If i have two 56-bit DES keys, and I use these two key to encrypt a 256-bit AES key. Like this: K_AT encrypts K_BT and K_BT encrypts K

What is the effective security of key K?

Is it 2 X 56 Bits?

Ohnana
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Ricky
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    This a homework question Ricky? Also, exactly what does `2X56 bits` mean? Are you saying that it is 112 bits? Something else? I think it is easier to work it out using exponents and then convert back to bits at the end. So your equation is `2X2^56`. That's math that I understand. – Neil Smithline Apr 18 '16 at 17:59
  • @Neil Smithline The homework question is easier than this one. This is a question that my teacher asked in lecture, just food for thought. Can you elaborate more? Like why is it 2x2^56? I thought the answer should be 2^112. I am still bit confused. Still, thank you!!!!! – Ricky Apr 19 '16 at 19:27
  • I wasn't saying that the math works out one way or the other, just trying to understand what you meant with the equation. – Neil Smithline Apr 19 '16 at 19:37
  • I'm not sure why this hasn't gotten more attention here Ricky. I'm going to start the process to have it moved to Crypto.SE. I think it will do better there. – Neil Smithline Apr 19 '16 at 19:37
  • BTW, I don't think this is more complicated than 2^56. An attacker could just forget about K_AT and brute force the unencrypted value of K_BT. There are only 2^56 possible values for that. They would use each guess to decrypt K and then test K by decrypting the cipher text. I think that works. – Neil Smithline Apr 19 '16 at 20:27
  • I'm not 100% certain given the notation used in the post, but I suspect what you're attempting to describe is double-DES, which has a strength of roughly `2^57` due to a meet-in-the-middle attack. If that in fact is what the problem is, see [these](http://security.stackexchange.com/a/42176/12) [existing](http://security.stackexchange.com/a/57061/12) answers for more details. – Xander Apr 19 '16 at 21:11
  • @Xande the link(postimg.org/image/nfbb6lck7) contains a full picture of the protocol. As the pictures shown, is it still 2^57 times instead of 2^112? Great helps and thank you! – Ricky Apr 19 '16 at 23:57
  • @NeilSmithline The link(postimg.org/image/nfbb6lck7) contains a full picture of the protocol. As the pictures shown, is it still 2^56 times instead of 2^112? Great helps and thank you! – Ricky Apr 19 '16 at 23:58

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