New algorithm for cube (third power) computation
zanoni at volterra.uniroma2.it
Fri Apr 9 11:52:15 CEST 2010
> Maybe it's easier to do something useful for fifth power? E.g., using
> (x^3 + 10) (x^2 + 5x + 10) = x^5 + 5 x^4 + 10 x^3 + 10 x^2 + 50 x + 100
> (x + 1)^5 = x^5 + 5 x^4 + 10 x^3 + 10 x^2 + 5x + 1
One should not forget the a_i and b_i coefficients: it seems that in this case
the problem is the final recomposition of longer coefficients than in the
I tryed to do some experiments with fifth power, but for what I have analyzed
it seems that recomposing (longer) coefficients a0^5, a1a0^5 etc. mixes
things up such in a way that the cube trick does not work any more (or at
least I was not able to see anything useful).
Anyway, I'll try again to have a look as soon as possible.
Centro Interdipartimentale "Vito Volterra"
Universita' degli Studi di Roma "Tor Vergata"
Via Columbia 2
00133 Roma, Italia
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