Re: What´s the algorithm that compresses a 20 digit big int, into 8 bytes ?
From: Bob Badour <bbadour_at_pei.sympatico.ca>
Date: Fri, 09 Apr 2010 14:20:27 -0300
Message-ID: <4bbf61e5$0$12418$9a566e8b_at_news.aliant.net>
> 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.
Date: Fri, 09 Apr 2010 14:20:27 -0300
Message-ID: <4bbf61e5$0$12418$9a566e8b_at_news.aliant.net>
>> 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. Received on Fri Apr 09 2010 - 19:20:27 CEST