On Sun, Dec 29, 2013 at 5:42 PM, meekerdb <meeke...@verizon.net> wrote:

>  On 12/29/2013 2:08 PM, Jason Resch wrote:
>
>
>
>
> On Sun, Dec 29, 2013 at 4:51 PM, meekerdb <meeke...@verizon.net> wrote:
>
>>  On 12/29/2013 1:28 PM, Jason Resch wrote:
>>
>>
>>
>>
>> On Sun, Dec 29, 2013 at 2:25 PM, meekerdb <meeke...@verizon.net> wrote:
>>
>>>  On 12/29/2013 5:56 AM, Bruno Marchal wrote:
>>>
>>>
>>>  On 28 Dec 2013, at 22:23, meekerdb wrote:
>>>
>>>  On 12/28/2013 4:09 AM, Bruno Marchal wrote:
>>>
>>> For a long time I got opponent saying that we cannot generate
>>> computationally a random number, and that is right, if we want generate
>>> only that numbers. but a simple counting algorithm generating all numbers,
>>> 0, 1, 2, .... 6999500235148668, ... generates all random finite
>>> incompressible strings,
>>>
>>>
>>> How can a finite string be incompressible?  6999500235148668 in base
>>> 6999500235148669 is just 10.
>>>
>>>
>>>
>>>  You can define a finite string as incompressible when the shorter
>>> combinators to generate it is as lengthy as the string itself.
>>> This definition is not universal for a finite amount of short sequences
>>> which indeed will depend of the language used (here combinators).
>>>
>>>  Then you can show that such a definition can be made universal by
>>> adding some constant, which will depend of the universal language.
>>>
>>>  It can be shown that most (finite!) numbers, written in any base, are
>>> random in that sense.
>>>
>>>  Of course, 10 is a sort of compression of any string X in some base,
>>> but if you allow change of base, you will need to send the base with the
>>> number in the message. If you fix the base, then indeed 10 will be a
>>> compression of that particular number base, for that language, and it is
>>> part of incompressibility theory that no definition exist working for all
>>> (small) numbers.
>>>
>>>
>>>  Since all finite numbers are small, I think this means the theory only
>>> holds in the limit.
>>>
>>> Brent
>>>
>>
>>
>>  Brent,
>>
>>  It is easy to see with the pigeon hole principal.  There are more 2
>> digit numbers than 1 digit numbers, and more 3 digit numbers than 2 digit
>> numbers, and so on.  For any string you can represent using a shorter
>> string, another "shorter string" must necessarily be displaced.  You can't
>> keep replacing things with shorter strings because there aren't enough of
>> them, so as a side-effect, every compression strategy must represent some
>> strings by larger ones.  In fact, the average size of all possible
>> compressed messages (with some upper-bound length n) can never be smaller
>> than the average size of all uncompressed messages.
>>
>>  The only reason compression algorithms are useful is because they are
>> tailored to represent some class of messages with shorter strings, while
>> making (the vast majority of) other messages slightly larger.
>>
>>
>>  A good explanation.
>>
>
>  Thanks.
>
>
>>  But just because you cannot compress all numbers of a given size doesn't
>> imply that any particular number is incompressible.
>>
>
>  That is true if you consider the size of the compression program to be
> of no relevance.  In such a case, you can of course have a number of very
> small strings map directly to very large ones.
>
>
>>   So isn't it the case that every finite number string is compressible in
>> some algorithm?  So there's no sense to saying 6999500235148668 is random,
>> but 11111111111111 is not, except relative to some given compression
>> algorithm.
>>
>
>  Right, but this leads to the concept of Kolmogorov complexity. If you
> consider the size of the minimum string and algorithm together, necessary
> to represent some number, you will find there are some patterns of data
> that are more compressible than others.  In your previous example with base
> 6999500235148668, you would need to include both that base, and the string
> "10" in order to encode 6999500235148669.
>
>
> But that seems to make the randomness of a number dependent on the base
> used to write it down? Did I have to write down "And this is in base 10" to
> show that 6999500235148668 is random?  There seems to be an equivocation
> here on "computing a number" and "computing a representation of a number".
>
>
>
A number containing regular patterns in some base, will also contain
regular patterns in some other base (even if they are not obvious to us),
compression algorithms are good at recognizing them.

The text of this sentence may not seem very redundant, but english text can
generally be compressed somewhere between 20% - 30% of its original size.
 If you convert a number like "55555555555" to base 2, its patterns should
be more evident in the pattern of bits.


>    For the majority of numbers, you will find the Kolmogorov complexity
> of the number to almost always be on the order of the number of digits in
> that number.  The exceptions like 1111111111 are few and far between.
>
>
> 111111111 looks a lot messier in base 9.
>
>

base 10: 1111111111111111111
base 9: 7355531854711617707
base 2: 11110110101101110101101010110010101111000100011100

In base 9, there is a high proportion of 7's compared to other digits. In
base 2, the sequence '110' seems more common than statistics would suggest.
 In any case, the number is far from incompressible. It takes only 9
characters to represent that 19 digit number in Kolmogorov complexity
"1r19inb10" = "1 repeated 19 times in base 10", in my encoding language.

Jason

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