At the moment, to get to K = 100, you are only using about 388 bits. So the size here is not all that large, only a few limbs.
In general arithmetic modulo primes that will fit in a limbs is generally faster than multiple precision arithmetic. But this is mainly due to there being less overhead. Some overhead is reintroduced in the CRT computation however, so the speedup may not be as huge as it might seem. How large is your N? Bill. On 17 May, 14:02, Tom Coates <t.coa...@imperial.ac.uk> wrote: > It turns out that for my problem it suffices to compute the > polynomials > > 1, f, f^2, ..., f^K > > modulo N for some large but known number N, so I suspect that it may > be faster to work modulo p for various primes p and then use the > Chinese Remainder Theorem. > > How large, roughly speaking, can I take my prime p to be without > slowing down the multiplication in (ZZ/pZZ)[x,y,z] ? > > Best, > > Tom > > -- > To post to this group, send an email to sage-devel@googlegroups.com > To unsubscribe from this group, send an email to > sage-devel+unsubscr...@googlegroups.com > For more options, visit this group athttp://groups.google.com/group/sage-devel > URL:http://www.sagemath.org -- To post to this group, send an email to sage-devel@googlegroups.com To unsubscribe from this group, send an email to sage-devel+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-devel URL: http://www.sagemath.org
