Shouldn't that be?
py> 16 * (-7/16 * math.log2(7/16) - 6/16 * math.log2(6/16)) =
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It looks like that averages my two examples. H by the way that equation is
really coolwhy does it return a high bit count when compared to >>>dec to
bin?
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Thomas Jollans
On 2017-10-25 23:22, danceswi...@gmail.com wrote:
> With every transform the entropy changes,
That's only true if the "transform" loses or adds information.
If it loses information, that's lossy compression, which is only useful
in very specific (but also extremely common)
So if the theoretical min compression limit (log2(n)*(x)) has a 3% margin but
your transform has a less than 3% inflate rate at most then there is room for
the transform to compress below the theoretical min. With every transform the
entropy changes, the potential for greater compression also
Whatever you do, you'll find that *on average* you
will need *at least* 34 bits to be able to represent
all possible 10-digit decimal numbers. Some might
be shorter, but then others will be longer, and
the average won't be less than 34.
The theoretical limit for arbitrary numbers 0 - 9 must
No leading zeroes are being dropped offwish this board has an edit button.
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Greg, you're very smart, but you are missing a big key. I'm not padding, you
are still thinking inside the box, and will never solve this by doing so. Yes!
At least you see my accomplishment, this will compress any random file.
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I did that quite a while ago. 352,954 kb.
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Finally figured out how to turn this into a random binary compression program.
Since my transform can compress more than dec to binary. Then i took a random
binary stream, changed it to a decimal stream 0-9 tranformed it into a
compressed/encrypted binary stream 23.7% smaller. Yes! Decode
Wow, do programmers actually use zscii. That is huge. So much wated space.
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Good point
I hope it has a use, other than a cute toyi don't see it yet.
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Thanks Paul...blunt to the point.
My 8 year old can decode this back into base 10, i still have to help him a bit
going from base 10 to 8 bit bytesit's incredibly simple to decode. No
dictionary, can easily be done with pencil and paper, does not rely on
redundancies.
Jon Hutton
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Just trying to find a practical application for this alg. Not real useful as it
stands now.
Jon Hutton
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I'm really not trolling, and even though some are sarcastic i sm learning from
your comments. Dec to bin is not bad at removing wasted space, but there is a
better way. Here is an example. How would you compress these numbers. If you
look for redundancy and then code to a bulky dictionary or
I really do not think this has a value besides being a trinket or cute toy.
Like i said i can not see how it can be adapted to work as a rand binary
compression alg...it only works with 0-9 in any seq. It's taken me six years to
solve, but so what.
Jon Hutton
danceswithnumb...@gmail.com
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According to this website. This is an uncompressable stream.
https://en.m.wikipedia.org/wiki/Incompressible_string
12344321
It only takes seven 8 bit bytes to represent this
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Ridiculous? Ludicrous??
Harsh words! First let me clarify before you lump this in with perpetual
motion, or cold fusion. It is a mapping solution to compress ANY i repeat ANY
random file with numbers of only 0 - 9 such as are in the million rand numbers
page. Entirely possible. Since i did it,
On Monday, July 11, 2016 at 11:52:27 AM UTC-6, jonas.t...@gmail.com wrote:
> What kind of statistic law or mathematical conjecture or is it even a
> physical law is violated by compression of random binary data?
>
> I only know that Shanon theorised it could not be done, but were there any
>
In Short, I cannot find a single mathematical proof that says you cannot
compress random numbers. Pigeon hole and other conjectures are just that. In
fact, the biggest fallacy when people start talking about compression is to say
that all compression alg rely on redundancies, or repetitive
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