Is floating point arbitrary precision available?

Question

Just for fun and because it was really easy, I’ve written a short program to generate Grafting numbers, but because of floating point precision issues it’s not finding some of the larger examples.

def isGrafting(a):
  for i in xrange(1, int(ceil(log10(a))) + 2):
    if a == floor((sqrt(a) * 10**(i-1)) % 10**int(ceil(log10(a)))):
      return 1

a = 0
while(1):
  if (isGrafting(a)):
    print "%d %.15f" % (a, sqrt(a))
  a += 1

This code misses at least one known Grafting number. 9999999998 => 99999.99998999999999949999999994999999999374999999912... It seems to drop extra precision after multiplying by 10**5.

>>> a = 9999999998
>>> sqrt(a)
99999.99999
>>> a == floor((sqrt(a) * 10**(5)) % 10**int(ceil(log10(a))))
False
>>> floor((sqrt(a) * 10**(5)) % 10**int(ceil(log10(a))))
9999999999.0
>>> print "%.15f" % sqrt(a)
99999.999989999996615
>>> print "%.15f" % (sqrt(a) * 10**5)
9999999999.000000000000000

So I wrote a short C++ program to see if it was my CPU truncating the floating point number or python somehow.

#include <cstdio>
#include <cmath>
#include <stdint.h>

int main()
{
  uint64_t a = 9999999998;
  printf("%ld %.15f %.15f %.15f %.15fn", a, sqrt((double)a), sqrt((double)a)*1e4, sqrt((double)a)*1e5, sqrt((double)a)*1e6);
  a = 999999999998;
  printf("%ld %.15f %.15f %.15f %.15fn", a, sqrt((double)a), sqrt((double)a)*1e5, sqrt((double)a)*1e6, sqrt((double)a)*1e7);
  a = 99999999999998;
  printf("%ld %.15f %.15f %.15f %.15fn", a, sqrt((double)a), sqrt((double)a)*1e6, sqrt((double)a)*1e7, sqrt((double)a)*1e8);
  return 0;
}

Which outputs:

9999999998 99999.999989999996615 999999999.899999976158142 9999999999.000000000000000 99999999990.000000000000000
999999999998 999999.999998999992386 99999999999.899993896484375 999999999999.000000000000000 9999999999990.000000000000000
99999999999998 9999999.999999899417162 9999999999999.900390625000000 99999999999999.000000000000000 999999999999990.000000000000000

So it looks like I’m running up hard against the limits of floating point precision and the CPU is chopping off the remaining bits because it thinks that the remaining difference is floating point error. Is there a way to work around this under Python? Or do I need to move to C and use GMP or something?

Asked By: OmnipotentEntity

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Answer 1

You can try with Decimal instead of floatingpoint.

Answered By: f p

Answer 2

Python has no built-in arbitrary-precision floats, but there are 3rd-party Python packages that use GMP: gmpy and PyGMP.

Answered By: Ned Batchelder

Answer 3

In the standard library, the decimal module may be what you’re looking for. Also, I have found mpmath to be quite helpful. The documentation has many great examples as well (unfortunately my office computer does not have mpmath installed; otherwise I would verify a few examples and post them).

One caveat about the decimal module, though. The module contains several in-built functions for simple mathematical operations (e.g. sqrt), but the results from these functions may not always match the corresponding function in math or other modules at higher precisions (although they may be more accurate). For example,

from decimal import *
import math

getcontext().prec = 30
num = Decimal(1) / Decimal(7)

print("   math.sqrt: {0}".format(Decimal(math.sqrt(num))))
print("decimal.sqrt: {0}".format(num.sqrt()))

In Python 3.2.3, this outputs the first two lines

   math.sqrt: 0.37796447300922719758631274089566431939601898193359375
decimal.sqrt: 0.377964473009227227214516536234
actual value: 0.3779644730092272272145165362341800608157513118689214

which as stated, isn’t exactly what you would expect, and you can see that the higher the precision, the less the results match. Note that the decimal module does have more accuracy in this example, since it more closely matches the actual value.

Answered By: Ricardo Altamirano

Answer 4

For this particular problem, decimal is a great way to go, because it stores the decimal digits as tuples!

>>> a = decimal.Decimal(9999999998)
>>> a.as_tuple()
DecimalTuple(sign=0, digits=(9, 9, 9, 9, 9, 9, 9, 9, 9, 8), exponent=0)

Since you’re looking for a property that is most naturally expressed in decimal notation, it’s a bit silly to use a binary representation. The wikipedia page you linked to didn’t indicate how many “non-grafting digits” may appear before the “grafting digits” begin, so this lets you specify:

>>> def isGrafting(dec, max_offset=5):
...     dec_digits = dec.as_tuple().digits
...     sqrt_digits = dec.sqrt().as_tuple().digits
...     windows = [sqrt_digits[o:o + len(dec_digits)] for o in range(max_offset)]
...     return dec_digits in windows
... 
>>> isGrafting(decimal.Decimal(9999999998))
True
>>> isGrafting(decimal.Decimal(77))
True

I think there’s a good chance the result of Decimal.sqrt() will be more accurate, at least for this, than the result of math.sqrt() because of the conversion between binary representation and decimal representation. Consider the following, for example:

>>> num = decimal.Decimal(1) / decimal.Decimal(7)
>>> decimal.Decimal(math.sqrt(num) ** 2) * 7
Decimal('0.9999999999999997501998194593')
>>> decimal.Decimal(num.sqrt() ** 2) * 7
Decimal('1.000000000000000000000000000')

Answered By: senderle

Answer 5

use decimal, (here is a clearer example):

>>> 2.3-2.2
0.09999999999999964
>>> from decimal import Decimal
>>> Decimal('2.3')-Decimal('2.2')
Decimal('0.1')
>>> float(Decimal('2.3')-Decimal('2.2'))
0.1
>>>

Answered By: U12-Forward

Is floating point arbitrary precision available?

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