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How to calculate the average key search time for the double des ?Uses standard DES with 56 key bits, and we can test 10^6 keys per second. Many thanks!

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Here's how you would go about this type of task:

  1. Find the most efficient attack on the cryptosystem under consideration.
  2. Find the time-complexity of this attack.
  3. Divide the time-complexity by the "computation power" you have.

As a hint: The time-complexity to break double DES is significantly smaller than $2^{112}$.

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    $\begingroup$ If the attacker use meet-in-the-middle attack. It only need about 2*2^56 ? $\endgroup$ – Edmund Ng Apr 5 '17 at 14:06
  • $\begingroup$ @EdmundNg yes, indeed. $\endgroup$ – SEJPM Apr 5 '17 at 14:08
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We know that in double DES the output of the first DES goes into the second DES as input and the output generated by the second DES function is the actual output.

At $10^6$ keys per second, we have $2^{56} \times 2^{56}$ possible keys for double DES brute force attack. That is we have $2^{112}$ possible keys.So it would take $\frac{2^{112}}{10^6}$ seconds or $1.6 \times 10^{20}$ years.The avg time would be half of that.

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    $\begingroup$ This is wrong. Double DES is vulnerable to MITM, so it's not $2^{56} \times 2^{56} = 2^{112}$ but $2^{56} + 2^{56} = 2^{57}$. $\endgroup$ – forest May 3 at 1:20

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