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  • CG 9:17 pm on March 30, 2013 Permalink | Reply
    Tags: arithmetic, , polynomial basis, reduction,   

    Polynomial Basis Squaring 

    Finally have successfully found some spare time to do coding to solve this polynomial squaring:

    OLYMPUS DIGITAL CAMERA

    And this is the result, x^5 + x + 1 :

    Screen shot 2013-03-30 at 8.58.52 PM

     
    • rudi 9:37 pm on May 30, 2013 Permalink | Reply

      bu, akan lebih cantik kalo nulis polinomnya pake latex,
      x^5 + x+1

      • CG 11:00 am on May 31, 2013 Permalink | Reply

        iya belum sempet dirapihin 😀 biasanya saya pake latex for wordpress

    • Akshay 11:18 pm on July 26, 2013 Permalink | Reply

      Hey please mail me this complete C code. sorry I’m not execute this ….please help me…please ..

  • CG 2:15 pm on February 7, 2009 Permalink | Reply
    Tags: , polynomial basis   

    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 😉

    [big headache continues…]

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

    Answer for #1 

    Again, from here:

    1. How to convert from polynomial to normal bases?

    Answer:
    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} .

     
  • CG 1:58 pm on February 2, 2009 Permalink | Reply
    Tags: , , polynomial basis   

    Answer for #2 

    Answering #2 from here:

    2. For what conditions ONB representation is available?

    Answer: One condition is when the p(x) generates GF elements which are linearly independent.

     
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