That actually doesn't make it "faster" as you claim. You are just
splitting the work out to N workers, which ignores big O notation
boundaries. A more appropriate sieve, could be the Euler sieve. However,
I am not sure what you mean by create a sieve that ignores the first few
prime numbers, could you expound on that?
On Tue, Dec 18, 2012 at 9:55 AM, Don <dondod...@gmail.com> wrote:
> He is how I did it:
>
> I used 48 parallel sieves, one for each value in the range 0..210
> which is not divisible by 2,3,5, or 7. If those numbers are store in
> A[48]={1,11,13.17,19,23,...}, then the Nth bit in sieve S represents
> the value N*210+A[S]. When a new prime P is encountered, you need to
> find the first place in each of the 48 sieves where it will be
> encountered. That will be the smallest value V > P^2 where V%210=A[S].
> Then to process a sieve you take steps of size 210*P. This is about
> four times faster than using a single sieve. It could be extended to
> include multiples of 11, but I didn't do that.
>
> Don
>
> On Dec 7, 4:14 pm, Don <dondod...@gmail.com> wrote:
> > I know that the Sieve of Eratosthenes is a fast way to find all prime
> > numbers in a given range.
> > I noticed that one implementation of a sieve spends a lot of time
> > marking multiples of small primes as composite. For example, it takes
> > 1000 times as long to mark off all of the multiples of five as it
> > takes to mark off the multiples of 5003. In addition, when it is
> > marking off the multiples of larger primes, most of them are multiples
> > of small primes. In fact, if it could skip over multiples of 2,3,5,7,
> > and 11, the sieve would be about 5 times faster.
> >
> > Can someone describe a way to make a sieve faster by not having to
> > deal with multiples of the first few prime numbers?
> >
> > Don
>
> --
>
>
>
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