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!

_________________________________
Making Money \$150 An Hour

More to do, more to do

1. Check polynomial representation for different polynomial with same order, and convert it to normal basis
2. Find out whether ONB Type I is convertible to ONB Type II or vice versa. (Intuitively, this can be done, hmmm…)
3. Think how to choose curve applicable for all those representations. Is it better find the polynomial then convert it to ONB or is it easier to do it the other way around?

Still in the middle writing the papers, now got stuck on choosing the curve that works with polynomial, ONB Type I and Type II.

For example, referring to Rosing’s book, the poly_prime 155bits is not really a prime polynomial, while the field size available for both Type I and II is 148bits. 155bits only for Type II.

Then after a curve has been choosed, I still have to make a code to convert that from polynomial basis to ONB I and II.

When will it be finished? Grrrrr 😦

How to compare?

Still trying to finish one paper and got stuck on how to compare the performance of one basis with another.

Space complexity? Time complexity? Execution time?

PhD to do

1. Dissertation focus: side channel attack
2. Write two (or maybe three) papers
3. Read Kumar’s book and contact him for further discussions
4. Find international journals

• Budi Rahardjo 3:06 pm on February 14, 2009 Permalink | Reply

Don’t forget to report (and document) progress (or lack of? 🙂 )

Discussions: help a lot

Now move on to plaintext embedding algorithms.

see…see… temen berantem as a sparring is greatly needed 😀

Basis conversion with matrix

Basis conversion can be done easier with matrix 🙂

• CG 4:27 pm on February 7, 2009 Permalink | Reply Tags: optimal normal basis ( 3 )

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 ..

From one basis to another

Apparently converting from one basis to another like from polynomial to normal basis is not as easy as I thought, hmm… have spent days scribbling, thinking, frustated, madly curious, and end up browsing several papers about that and being succesfully diverted from the main target of doing paper on plaintext embedding, oh my!

Let me digest some more papers I have just downloaded 30secs ago, and will post something useful here as soon as possible 😉

• CG 2:38 pm on February 2, 2009 Permalink | Reply Tags: normal basis ( 4 ), polynomial basis ( 4 )

Again, from here:

1. How to convert from polynomial to normal bases?

GF element can be represented in Polynomial Basis (PB) or Normal Basis (NB).

For example we have polynomial $p(x)=x^{3}+x^{2}+1$

The PB representation in $GF\left (2^{3} \right )$ is

If $GF\left (p^{m} \right )$ be a field with $p^{m}$ elements and $\beta$ an element of it such that $m$ elements $\left \{\beta ,\beta ^{p}, ... , \beta ^{p^{m-1}} \right \}$ are linearly dependent. Then this set forms a normal basis for $GF\left (p^{m} \right )$

The NB representation of elements in $GF\left (2^{3} \right )$ will only use 3 elemens $\beta$, $\beta ^{2}$ dan $\beta ^{4}$.

c
Compose new post
j
Next post/Next comment
k
Previous post/Previous comment
r