Understanding Cryptography by Christof Paar and Jan Pelzl - Chapter 4 Solutions - Ex4.6

- 2 mins

Exercise 4.6

Compute in :

where the irreducible polynomial is the one used by AES, .

Note that Table 4.2 contains a list of all multiplicative inverses for this field.


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.

The multiplicative inverse could be found via the Euclidian Algorithm, though in this instance I have simply looked it up in the table mentioned above:

Next we perform a naive multiplication of with the inverse we just looked up:

All that’s left to do now is to reduce it via the reduction polynomial

The result of this calculation is therefore . This answer can be verified as correct (assuming my code is correct) by placing the following python code in the __main__ section of the script written for Ex4.3:

    mod = Mod2Polynomial([1, 1, 0, 1, 1, 0, 0, 0, 1])
    a = Mod2Polynomial([1, 1, 0, 0, 1])
    b = Mod2Polynomial([1, 1, 0, 1, 1])

    print a * b % mod

Thomas Busby

Thomas Busby

I write about computing stuff

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