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alexdok [17]
3 years ago
11

Assume that you are able to do an exhaustive search for the key to an encrypted message at the rate of 100 Million trials per se

cond. (a) (7 Points) How long in (days or years) would it take for you to find the key on the average to a code that used 112-bit key on average
Computers and Technology
1 answer:
pychu [463]3 years ago
5 0

Answer:

8.22 × 10²⁰ years

Explanation:

Given that:

Total frequency = 100 million per record

The length of the key used for the encryption = 112 bit key

To calculate the number of seconds in  a year, we have:

= 365 × 24 × 60 × 60

= 3.1536 × 10⁷ seconds

Thus, on average, the number of possible keys that is required to check for the decryption should be at least  2¹¹¹ keys.

\mathbf{ = \dfrac{2^{111} \times 10^6}{3.1536 \times 10^7} = 8.22 \times 10^{20} \ years}

Thus, it will take a total time of about 8.22 × 10²⁰ years on average.

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2 years ago
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vova2212 [387]

The answer is Every six months

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3 years ago
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If the list [8,1,4,2,10,0] is sorted with the selection sort algorithm, what is the list at the 4th step of the algorithm?
Savatey [412]

Answer:

The answer to this question is given below in the explanation section.

Explanation:

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The correct answer after selection sort algorithm at the 4th step is:D

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The complete detail of step by step execution is shown in the attached document file.

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