Multiplication/Division in Galois Field (2^8)

Question

I'm attempting to implement multiplication and division in $GF(2^8)$ using log and exponential tables. I'm using the exponent of 3 as my generator, using instructions from here.

However I'm having trouble getting the expected answer for some of these trivial multiplications.

For $2 · 4$ this works:

$$ \begin{align*} \log_3(2) &= 25 \\ \log_3(4) &= 50 \\ 25 + 50 &= 75 \\ \exp_3(75) &= 8 \Rightarrow \text{ expected answer} \end{align*}$$

However for $7 · 11$ this breaks down:

$$\begin{align*} \log_3(7) &= 198 \\ \log_3(11) &= 104 \\ 198 + 104 &= 302. \\ \text{Mod it by 255, gives us 47.}\\ \exp_3(47) &= 49 \end{align*}$$ instead of the expected 77.

From what I understand, we use modulus 255 because the 3 generator 'wraps around' on the 255th power (so the pattern repeats after that) thus we only need 0~254. Even if I'm wrong in this, $302 \bmod 256$ still doesn't give us $70$, where $\exp_3(70) = 77$ (the expected answer)

My observation for both multiplication and division is that it works fine until the result of addition/subtraction goes out of range of 0~255.

Answer 1

In GF(2⁸), 7 × 11 = 49. The discrete logarithm trick works just fine.

Your mistake is in assuming that Galois field multiplication works the same way as normal integer multiplication. In prime-order fields this actually is more or less the case, except that you need to reduce the result modulo the order of the field, but in fields of non-prime order the multiplication rules are different.

Let's do 7 × 11, for example. Noting that 7 = 111₂ and 11 = 1011₂, we can calculate 7 × 11 in binary as:

    111 ×
   1011 =
   ------
    111 +
   1110 +
 111000 =
 --------

So far, everything works the same as in ordinary integer multiplication. But whereas in the integers we'd propagate carries while doing the addition, and thus end up with

      111₂ + 1110₂ + 111000₂ = 10101₂ + 111000₂ = 1001101₂ = 77,

in GF(2ⁿ) addition is done bitwise without carries (i.e. GF(2ⁿ) addition is the same as bitwise XOR), and thus we get

      111₂ + 1110₂ + 111000₂ = 1001₂ + 111000₂ = 110001₂ = 49.

(Of course, if the result exceeded the group order, we'd also have to reduce it modulo the reduction polynomial, but in this case that doesn't happen for n ≥ 7.)