I understand how run length codes work. If you want to know more about data
compression, maybe you will be interested in my book.
http://mattmahoney.net/dc/dce.html

To compress a unary representation of Graham's number G as a run length
code, you need to encode a binary representation of G. The only way you are
going to do that is to write a program or a mathematical description of G
because G itself has too many digits to write.

Maybe you don't understand how really big G is. The number of digits in G
is essentially G. The number of digits in that number is essentially G. The
number of digits in THAT number is essentially G. The number of times you
have to reduce it to where you could write it down is essentially G. You
really aren't getting anywhere trying to describe it in terms of numbers
that the human brain can comprehend. I mean nobody can comprehend how big
towers of exponents whose height is itself a tower of exponents like 3^^^3
is, or how vastly bigger 3^^^^3 is than 3^^^3. What hope do you have of
understanding G?


On Sat, Nov 2, 2019, 11:13 PM <immortal.discover...@gmail.com> wrote:

> Such a massive number as you mentioned can be compressed because there is
> a pattern in that bitstream or in the creation of that bitstream. The code
> 1111111111 is ten bits long yet it says 1024 of something (the pattern).
>
> You can store 4,294,967,296 'A's using basically 4 bytes. Here, the
> resulting huge number clearly has a pattern.
>
> You can store the bitstream '00' (3) along with an iteration count of 6
> (code 11), and add half of that (or less than half if uneven number). 00
> (3), 01 (4), 11 (6), 010 (9), 110 (13), 0100 (19), 1101 (28), 01011 (42).
> Here, the resulting huge number doesn't show a pattern. 10 bits stored can
> run this over 1024 times, and as we seen, 3, 4, 6, 9, 13, 19, 28, 42, 63,
> 94, 141, 211, ... meaning after just ~10 iterations we already have ~8 bits
> in length of a generated number.
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