# Understanding Cryptography by Christof Paar and Jan Pelzl - Chapter 4 Solutions - Ex4.13

- 4 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 4.13

We consider the first part of the ByteSub operation, i.e, the Galois field inversion.

- Using Table 4.2, what is the inverse of the bytes 29, F3 and 01, where each byte is given in hexadecimal notation?
- Verify your answer by performing a \(GF(2^8)\) multiplication with your answer and the input byte. Note that you have to represent each byte first as polynomials in \(GF(2^8)\). The most significant byte of each byte represents the \(x^7\) coefficient

### Solution

*This solution is verified as correct by the official Solutions for Odd-Numbered Questions manual.*

1. As per Table 4.2:

\[26_{16}^{-1} = A8_{16}\] \[F3_{16}^{-1} = 34_{16}\] \[01_{16}^{-1} = 01_{16}\]2. Instead of doing long-hand multiplications and reductions, I will be using the Mod2Polynomial class I wrote for earlier exercises.

Just as a reminder, the reduction primitive polynomial is the AES polynomial: \(P(x) = x^8 + x^4 + x^3 + x + 1\).

This is expressed as a Mod2Polynomial class as follows (it takes least significant bit first):

The first pair, \(26_{16}^{-1} = A8_{16}\), can be checked as follows:

\[26_{16} \equiv 00100110_2 \equiv x^5 + x^2 + x\] \[A8_{16} \equiv 10101000_2 \equiv x^7 + x^5 + x^3\]When multiplied together, these should produce an output of 1:

This does produce an output of 1.

The second pair, \(F3_{16}^{-1} = 34_{16}\), can be checked as follows:

\[F3_{16} \equiv 11110011_2 \equiv x^7 + x^6 + x^5 + x^4 + x + 1\] \[34_{16} \equiv 00110100_2 \equiv x^5 + x^4 + x^2\]When multiplied together, these should produce an output of 1:

This does produce an output of 1.

The 01 case is pretty trivial and doesn’t really need to be checked, but we will anyway:

\[01_{16} \equiv 00000001_2 \equiv 1\]When multiplied together, these should produce an output of 1:

This does produce an output of 1.