Tnx! I learned useful stuff ...
On 31 jan, 20:52, William Stein <[email protected]> wrote:
> On Sat, Jan 31, 2009 at 11:40 AM, William Stein <[email protected]> wrote:
> > On Sat, Jan 31, 2009 at 11:26 AM, mabshoff
> > <[email protected]> wrote:
>
> >> On Jan 31, 11:17 am, William Stein <[email protected]> wrote:
> >>> On Sat, Jan 31, 2009 at 11:15 AM, Rolandb <[email protected]> wrote:
>
> >>> > Hi,
>
> >>> > I received first a MemoryError, and later on Sage reported:
>
> >>> That means that Sage ran out of memory. It's not a bug -- it's just a
> >>> limitation of your computer.
>
> >> srange() builds a list in memory, so using an iterator will make that
> >> code use less memory.
>
> > Good idea. His an iterator version of the code:
>
> > uitkomst1=[]
> > uitkomst2=[]
> > eind=int((10^9+2)/(2*sqrt(3)))
> > print eind
> > for y in (1..eind):
> > test1=is_square(3*y^2+1,True)
> > test2=is_square(48*y^2+1,True)
> > if test1[0] and test1[1]%3==2: uitkomst1.append((y,(2*test1[1]-1)/3))
> > if test2[0] and test2[1]%3==1: uitkomst2.append((y,(2*test2[1]+1)/3))
> > print uitkomst1
> > een=sum(3*x-1 for (y,x) in uitkomst1 if 3*x-1<10^9)
> > print uitkomst2
> > twee=sum(3*x+1 for (y,x) in uitkomst2 if 3*x+1<10^9)
> > print een+twee
>
> > It prints
>
> > 288675135
> > ... and I got bored waiting for the rest...
>
> OK, I got really tired of waiting, and just wrote a cython version of
> the above, which does it for 10^9 in 25 seconds. It's over 200 times
> faster than your code. See attached worksheet or paste this into a
> notebook cell and press shift enter, then do f(10^9):
>
> %cython
> from sage.all import is_square
> cdef extern from "math.h":
> long double sqrtl(long double)
>
> def f(n):
> uitkomst1=[]
> uitkomst2=[]
> cdef long long eind=int((n+2)/(2*sqrt(3)))
> cdef long long y, t
> print eind
> for y in range(1,eind):
> t = <long long>sqrtl(<long long> (3*y*y + 1))
> if t * t == 3*y*y + 1:
> uitkomst1.append((y, (2*t-1)/3))
> t = <long long>sqrtl(<long long> (48*y*y + 1))
> if t * t == 48*y*y + 1:
> uitkomst2.append((y, (2*t+1)/3))
> print uitkomst1
> een=sum([3*x-1 for (y,x) in uitkomst1 if 3*x-1<10^9])
> print uitkomst2
> twee=sum([3*x+1 for (y,x) in uitkomst2 if 3*x+1<10^9])
> print een+twee
>
> I don't guarantee there isn't some funny overflow or wrongness with
> the above. All I claim is that it is faster than you'll get use any
> interprter.
>
> William
>
> sage_support_Rolandb.sws
> 27KWeergevenDownloaden
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