Re: What´s the algorithm that compresses a 20 digit big int, into 8 bytes ?

From: paul c <toledobythesea_at_oohay.ac>
Date: Fri, 09 Apr 2010 17:28:48 GMT
Message-ID: <ktJvn.1318$Z6.26_at_edtnps82>


paul c wrote:
> Bob Badour wrote:

>> paul c wrote:
>>
>>> Rafael Anschau wrote:
>>>
>>>> On Apr 9, 11:45 am, Bob Badour <bbad..._at_pei.sympatico.ca> wrote:
>>>>
>>>>> If it support the full 20 decimal digit range, no algorithm will 
>>>>> fit it
>>>>> into 64 bits so choosing a different algorithm will achieve nothing.
>>>>
>>>> Any proof of that, or is it just another hypothesis ? My intuition
>>>> tells me this is true, but I would like to see a proof of that.
>>>>
>>> Open up your calculator and look at the result of 2 to the power of 
>>> 63.  I get 9,223,372,036,854,775,808.  Only nineteen digits.
>>
>> But in my calculator (2^64)-1 gives 18,446,744,073,709,551,615 which 
>> has 20 digits.

>
> I'm sure it does. I didn't bother with complement nuances because the
> number of digits proves the point.

Sorry, meant to say the lack of high-order 2, 3, etc. proves the point. Received on Fri Apr 09 2010 - 19:28:48 CEST

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