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New to Py 3.3.3 having prob. with large integer div. float.


On Tuesday, February 11, 2014 2:24:01 AM UTC-5, cas... at gmail.com wrote:
> On Monday, February 10, 2014 6:40:03 PM UTC-8, hlauk.h... at gmail.com wrote:
> 
> > I am coming off Python 2.6.6 32 bite platform and now in win 8.1 64 bite.
> 
> > I had no problems with gmpy in 2.6.6 and large integer floating points where
> 
> > you could control the length of the floating point by entering the bite size
> 
> > of the divisor(s) you are using. That would give the limit length of the float
> 
> > in the correct number of bites.
> 
> > 
> 
> > In Python 3.3.3 and gmpy2 I have tried many things in the import mpfr module
> 
> > changing and trying all kinds of parameters in the gmpy2 set_context module 
> 
> > and others.
> 
> > 
> 
> > The best I can get without an error is the results of a large integer 
> 
> > division is a/b = inf. or an integer rounded up or down.
> 
> > I can't seem to find the right settings for the limit of the remainders in the
> 
> > quotient.  
> 
> > 
> 
> > My old code in the first few lines of 2.6.6 worked great and looked like this -
> 
> > 
> 
> > import gmpy
> 
> > 
> 
> > BIG =(A large composite with 2048 bites) 
> 
> > SML =(a smaller divisor with 1024 bites)
> 
> > 
> 
> > Y= gmpy.mpz(1)
> 
> > A= gmpy.mpf(1)
> 
> > 
> 
> > y=Y
> 
> > 
> 
> > x=BIG
> 
> > z=SML
> 
> > a=A
> 
> > k=BIG
> 
> > j=BIG
> 
> > x=+ gmpy.next.prime(x)
> 
> > 
> 
> > while y < 20: 
> 
> >     B = gmpy.mpf(x.1024)
> 
> > ## the above set the limit of z/b float (see below) to 1024   
> 
> >     b=B
> 
> >     a=z/b
> 
> >     c=int(a)
> 
> >     d=a-c
> 
> >     if d = <.00000000000000000000000000000000001:
> 
> >          proc. continues from here with desired results.
> 
> > 
> 
> > gmpy2 seems a lot more complex but I am sure there is a work around.
> 
> > I am not interested in the mod function.
> 
> > 
> 
> > My new conversion proc. is full of ## tags on the different things
> 
> > I tried that didn't work.
> 
> > 
> 
> > TIA 
> 
> > Dan
> 
> 
> 
> The following example will divide two integers with a result precision
> 
> of 1024 bits:
> 
> 
> 
> import gmpy2
> 
> 
> 
> # Set mpfr precision to 1024
> 
> gmpy2.get_context().precision=1024
> 
> 
> 
> # Omitting code....
> 
> 
> 
> a = gmpy2.mpz(SML)/gmpy2.mpz(x)
> 
> 
> 
> Python 3.x performs true division by default. When integer division involves
> 
> an mpz, the result will be an mpfr with the precision of the current context.
> 
> 
> 
> Does this help?
> 
> 
> 
> casevh

Thanks a lot, I will give it a go.

Dan