On Nov 7, 2013, at 21:25, jonas.thornv...@gmail.com wrote:

> Den fredagen den 8:e november 2013 kl. 03:17:36 UTC+1 skrev Chris Angelico:
>> On Fri, Nov 8, 2013 at 1:05 PM,  <jonas.thornv...@gmail.com> wrote:
>> 
>>> I guess what matter is how fast an algorithm can encode and decode a big 
>>> number, at least if you want to use it for very big sets of random data, or 
>>> losless video compression?
>> 
>> 
>> 
>> I don't care how fast. I care about the laws of physics :) You can't
>> 
>> stuff more data into less space without losing some of it.
>> 
>> 
>> 
>> Also, please lose Google Groups, or check out what other people have
>> 
>> said about making it less obnoxious.
>> 
>> 
>> 
>> ChrisA
> 
> Please, you are he obnoxious, so fuck off or go learn about reformulation of 
> problems. Every number has an infinite number of arithmetical solutions. So 
> every number do has a shortest arithmetical encoding. And that is not the 
> hard part to figure out, the hard part is to find a generic arithmetic 
> encoding.
> 
> I am not sure if it is just stupidness or laziness that prevent you from 
> seeing that 4^8=65536.

Chris's point is more subtle: the typical computer will store the number 65536 
in a single byte, but it will also store 4 and 8 in one byte. So if your choice 
is between sending "65536" and "(4,8)", you actually loose efficiency in your 
scheme. Don't think in decimal, but in terms of information needing transfer. 

You might reply that you don't need a whole byte for 4 or 8 -- that's true. You 
could, e.g., just encode the fourth and eight bits of a single byte and send 
that: certainly gives some compression but at the cost of generality-- what 
would you do with 65^65? It's sort of like the mean-variance tradeoff in 
statistics; it's much easier to encode certain data sets (e.g. powers of two as 
you noted) but only if you concede your algorithm will only work for those 
values. 

More generally, check out the work of Claud Shannon; a very accessible and 
worthwhile author. 
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