0. One of these numbers needs to have an x to get extended precision.
1. 28433 * 2^7830457 + 1
is the same as
28433 * 2^7830458
which seems unlikely to be what was meant.
2. Roger, isn't the power done by repeated squaring or some such trick?
So the power will take about 30-50 multiplications whose average
length^2 is around 1500000^2, right? That could give an estimate
of the expected time.
Henry Rich
---- Roger Hui <[email protected]> wrote:
> The multiplication is currently O(n^2) so the number
> you described will take a very long time. How long?
> You can gauge it by timing the multiplication of
> numbers with 1e4, 2e4, 4e4, etc. digits.
>
> The multiplication would be faster with:
> http://www.jsoftware.com/jwiki/Essays/FFT
>
>
>
> ----- Original Message -----
> From: David Vaughan <[email protected]>
> Date: Wednesday, July 6, 2011 8:32
> Subject: [Jprogramming] Large numbers
> To: Programming forum <[email protected]>
>
> > Hi, what's the largest number J can compute?
> > I'm looking to compute 28433 * 2^7830457 + 1, which has 2357207
> > digits. I started calculating it, but it's taking a very long
> > time (obviously), an I'm wondering how long it's likely to take,
> > and whether it's even possible.
>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
