That is: no compression, but it won't make things worse either.
Unless the input data is the digits of pi, obviously, or the result of some computation involving pi.
You can use all the math stuff like scientific notation, tetration, etc... but it won't help you make things smaller.
Math notation is a form of compression. 10^9 is 1000000000, compressed. But the offset into pi is effectively a random number, and you can't compress random numbers no matter what technique you use, including math notation.
This can be formalized and mathematically proven. The only thing wrong here is that pi is not a random number, but unless you are dealing with circles, it looks a lot like it, so while unproven, I think it is a reasonable shortcut.
It is meant to be a joke anyway.
Most algs that I have ever made. There are several places where your gains disappear. The dictionary lookup for me is where things come apart. Sometimes it is the encoding of the bytes/blocks themselves.
In your example you could find all of the possible 8k blocks out there in pi. Now that number set would be very large. So it will be tough to get into your head how it is working. As it is not the whole of pi space you also probably need a dictionary or function to hold it or at least pointers to it.
One way to tell if a compression alg is doing ok is to try to make the most minimal version of it then scale it out. For example start with a 4 bit/8 bit/16 bit value instead of 8k. Then see how much space it would take up. Now sometimes scaling it up will let you get better gains (not always). That is where you will have a pretty good idea if it works or not. Like just move from 1 byte to 2 then 4 and so on. Just to see if the alg works. That exercise also lets you see if there are different ways to encode the data that may help as well.
I got nerd sniped about 3 decades ago on problems just like this. Still trying :)
I guess when God made the Universe, he filled it with CSAM, hate speech and unsolicited Viagra ads...
[1]https://e.math.cornell.edu/people/Nikhil_Sahoo/files/pi_pape...
rot13 spoiler
gur raqvat bs Pbagnpg juvpu vf vzcbffvoyr
rot13For example,
pi = the square root of the sum of 6/x^2 over x = (1, 2, 3...).
pi = the smallest positive value of ln(-1)/i.
pi = half the period of non-zero solutions to the differential equation f = -f''.
On a more serious note, as far as I understand these compression competitions require that static data is included in the size computation. So if you compress 1000 MB into 500 MB, but to decompress you need a 1 MB binary and a 100 MB initial dictionary, your score would be 500 + 100 + 1 = 601 MB, not 500 MB.
The relevance to this discussion is that the LLM weights would have to be included as static data, since the only way to regenerate them is from the initial training data, which is much larger than the resulting model. By comparison, pi based compression is the other way around: since pi is a natural constant, if your decompressor requires (say) a trillion digits of pi, you could write a relatively small program (a few kb) to generate them. It would be terribly slow, but it wouldn't affect your compression ratio much.
Well, either your program 'works', or you will have discovered a major new insight about Pi.
> On a more serious note, as far as I understand these compression competitions require that static data is included in the size computation. So if you compress 1000 MB into 500 MB, but to decompress you need a 1 MB binary and a 100 MB initial dictionary, your score would be 500 + 100 + 1 = 601 MB, not 500 MB.
And that's the only way to do this fairly, if you are running a competition where you only have a single static corpus to compress.
It would be more interesting and would make the results more useful, if the texts to be compressed would be drawn from a wide probability distribution, and then we scored people on eg the average length. Then you wouldn't necessarily need to include the size of the compressor and decompressor in the score.
Of course, it would be utterly impractical to sample Gigabytes of new text each time you need to run the benchmark: humans are expensive writers. The only way this could work would be either to sample via an LLM, but that's somewhat circular and wouldn't measure what you actually want to measure in the benchmark, or you could try to keep the benchmark text secret, but that has its own problems.
Ha, next: a compression algorithm that requires the user to first build an infinite library...
> a compression algorithm that requires the user to first build an infinite library...
Kind of already exists, pifs. More like a joke, but the concept is already a joke so...
Fascinating. Do you know if this has been proven about any interesting number (that wasn't explicitly constructed to make this true)?
Now an interesting problem would be to have the most “compact” number, where you can find all the strings of length n in a minimal number of digits of such number. It reminds me of the math problem solved by a 4chan user.