Choosing n and m for composite field

Referring to “Efficient Normal Basis Multipliers in Composite Fields” – Sangho Oh, Chang Han Kim, Jongin Lim, and Dong Hyeon Cheon, there is classification of hardware-applicable composite fields:

1. Type I composite field where a subfield $GF(2^n)$ in ONB2 and an extension field $GF(2^{nm})$ in ONB1
2. Type II composite field where a subfield $GF(2^n)$ in ONB1 and an extension field $GF(2^{nm})$ in ONB2
3. Type III composite field where a subfield $GF(2^n)$ in ONB2 and an extension field $GF(2^{nm})$ in ONB2

This is different with composite fields presented in “Efficient Methods for Composite Field Arithmetic” – E. Sava ̧s and C ̧. K. Koc, where the selection of $n$ and $m$  does not put their normal basis types (ONB1 or ONB2) into consideration.

Now the questions are:

1. Would it be better if we choose $n$, $m$ and $nm$ in ONB1/ONB2?
2. Which polynomial irreducible to be used? With degree = $n$, or degree = $m$ or degree = $nm$?

[pounding headache, and without answering these questions i wouldnt be able to start the hw design.]

ONB Type II, why?

Why do we need to use two field elements from two different field to generate a Type II ONB?

First pick an element $\gamma$ of order $2m+1$ in $F_{2}^{2m}$ to find $\beta$ in field $F_{2}^{m}$.

Why?

Just passing by.Btw, your website have great content!

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• CG 4:27 pm on February 7, 2009 Permalink | Reply Tags: optimal normal basis

Answering #2 from list of questions here:

An ONB of Type I exists a given field $GF(2^{m})$ if:

• $m+1$ is a prime
• 2 is a primitive in $GF(m+1)$

A Type II optimal normal basis exists in $GF(2^{m})$ if:

• $2m+1$ is prime
• either 2 is a primitive in $GF(2m+1)$ or $2m+1\equiv 3 \left (mod\; 4 \right )$ and 2 generates the quadratic residues in $GF(2m+1)$

Interesting notes:

An ONB exists in $GF(2^{m})$ for 23% of all possible values of $m$

said this paper. Hmmm, that’s something.

hmm, gak ngerti ..

c
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