# Understanding Cryptography by Christof Paar and Jan Pelzl - Chapter 4 Solutions - Ex4.16

- 2 mins- Return to index
- Exercise 4.1
- Exercise 4.2
- Exercise 4.3
- Exercise 4.4
- Exercise 4.5
- Exercise 4.6
- Exercise 4.7
- Exercise 4.8
- Exercise 4.9
- Exercise 4.10
- Exercise 4.11
- Exercise 4.12
- Exercise 4.13
- Exercise 4.14
- Exercise 4.15
- Exercise 4.16

## Exercise 4.16

For the following, we assume AES with 192-bit key length. Furthermore, let us assume an ASIC which can check \(3 \times 10^7\) keys per second.

- If we use 100,000 such ICs in parallel, how long does an average key search take? Compare this period of time with the age of the universe (approx. \(10^{10}\) years).
- Assume Moore’s Law will still be valid for the next few years, how many years do we have to wait until we can build a key search machine to perform an average key search of AES-192 in 24 hours? Again, assume that we use 100,000 ICs in parallel.

### Solution

*I haven’t yet verified this solution independently. If you spot any mistakes, please leave a comment in the Disqus box at the bottom of the page.*

1. With 100,000 ICs in parallel, the amount of keys we can check per second is as follows:

\[10^5 \,\mathsf{ICs} \times 3 \times 10^7\,\mathsf{keys/second} = 3 \times 10^{12}\,\mathsf{keys/second}\]The size of the keyspace in this instance is \(2^{192}\), meaning the average key search will need to do \(2^{192} \div 2 = 2^{191}\) checks. As such, the time required for an average key search is as follows:

\[2^{191} \,\mathsf{keys} \div 3 \times 10^{12}\,\mathsf{keys/second} \approx 1.046 \times 10^{45} \,\mathsf{seconds}\]We can calculate what this value is in years:

\[\frac{1.046 \times 10^{45} \,\mathsf{seconds}}{60 \times 60 \times 24 \times 365.25} \approx 3.315 \times 10^{37} \,\mathsf{years}\]This is approximately \(10^{28}\) times the current elapsed age of the universe.

2. If we represent the number of Moore’s Law iterations to bring the using \(i\) then we can express the equation as:

\[5.304 \times 10^{38}\,\mathsf{years} \times \frac{365.25}{2^i} = 1\,\mathsf{day}\]We can rearrange this equation to make \(2^i\) the only item on the left:

\[2^i = 5.304 \times 10^{38}\,\mathsf{years} \times 365.25\,\mathsf{days}\]The reveals a value of \(i\) which is roughly \(133.153\).

Rounding the number of iterations to 134 allows us to calculate the number of years:

\[1.5\,\mathsf{years} \times 134\,\mathsf{iterations} = 201\,\mathsf{years}\]