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If we have a symmetric algorithm with key length $m$ bit, and we do a 5 time encryption with it. How much will it be its effective key length.

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  • $\begingroup$ I don't get that solution; what does it cost to brute force the key in your opinion? Besides that, I don't know how good your math is. Mine is pretty bad, but that solution, sheesh. $\endgroup$ – Maarten Bodewes Jun 18 at 23:03
  • $\begingroup$ Yes. For example in book i read about Triple Des There are 22k possibilities to run through all possible keys of two encryptions or decryptions. In the case of 3DES, this forces an attacker to perform 2^112 key tests, which is entirely infeasible with current technology. In summary, the meet-in-themiddle attack reduces the effective key length of triple encryption from 3κ to 2κ. Because of this, it is often said that the effective key length of triple DES is 112 bits as opposed to 3 · 56 = 168 bits which are actually used as input to the cipher. $\endgroup$ – MrJab Jun 18 at 23:05
  • $\begingroup$ What makes 3 key triple DES and 2 key triple DES different? What about one-key triple DES? Or, from an entirely different perspective, what if I extended the number of rounds of a cipher by a factor 3? $\endgroup$ – Maarten Bodewes Jun 18 at 23:08
  • $\begingroup$ Final hint, using my mod powers :) How many keys are you using in your question? How is brute force, i.e. trying all the keys affected in that case? $\endgroup$ – Maarten Bodewes Jun 18 at 23:13
  • $\begingroup$ Hmm, in hindsight that question doesn't specify if 5 different keys are used or not. $\endgroup$ – Maarten Bodewes Jun 18 at 23:17

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