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From: Dennis Lee Bieber <wlfraed@ix.netcom.com>
Newsgroups: comp.lang.python
Subject: Re: Serious error in int() function?
Date: Thu, 14 Apr 2016 20:36:12 -0400
Organization: IISS Elusive Unicorn
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On Thu, 14 Apr 2016 13:07:03 +0100, blindanagram@nowhere.net declaimed the
following:

>On 14/04/2016 09:13, blindanagram@nowhere.net wrote:
>> On 14/04/2016 08:59, blindanagram@nowhere.net wrote:
>>> On 14/04/2016 07:52, ast wrote:
>> 
>>> This means that the result will be correct provided it has 53 or less
>>> bits - just short of 16 decimal digits (i.e for square numbers with less
>>> than 32 digits).
>>>
>>> With an integer square root function (isqrt), this program:
>>>
>>> for exp in count(20):
>>>   v = 2 ** exp - 1
>>>   if isqrt(v) != sqrt(v):
>>>     print(exp, isqrt(v), sqrt(v))
>>>     break
>>>
>>> terminates with:
>>>
>>>   54 18014398509481983 1.8014398509481982e+16
>>>
>>> showing a first error for a 54 bit square root
>>>
>> I should also have said that the square root of integer squares with
>> between 15 and 30 decimal digits will only be correct if the square
>> numbers themselves are exactly representable in 53 bits.  So we can
>> expect failures for squares with 16 or more digits.
>
>However, if a number with 31 or less digits is known to be the square of
>an integer, the IEEE754 sqrt function will (I believe) give the correct
>result.

	How could it EXCEPT by having ~15 significant digits and an exponent --
since that is all the data that is provided by a double precision floating
point. That is, for example, 

>>> 1000000000000000.0 * 1000000000000000.0
1e+30
>>> import math
>>> math.sqrt(1e30)
1000000000000000.0
>>> 

only has ONE significant digit -- even though it has thirty 0s before the
decimal point.

>>> 100000000000000000.0 * 100000000000000000.0
1e+34
>>> math.sqrt(1e34)
1e+17
>>> 10000000000000000.0 * 10000000000000000.0
1e+32
>>> math.sqrt(1e32)
1e+16
>>> 

-- 
	Wulfraed                 Dennis Lee Bieber         AF6VN
    wlfraed@ix.netcom.com    HTTP://wlfraed.home.netcom.com/