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$Z6.26_at_edtnps82>
>
> I'm sure it does. I didn't bother with complement nuances because the
> number of digits proves the point.
Date: Fri, 09 Apr 2010 17:28:48 GMT
Message-ID: <ktJvn.1318$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