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Fabrice Bellard's TS Zip (2024)

230 points1 daybellard.org
omoikane1 day ago

Current leader of the Large Text Compression Benchmark is NNCP (compression using neural networks), also by Fabrice Bellard:

https://bellard.org/nncp/

Also, nncp-2024-06-05.tar.gz is just 1180969 bytes, unlike ts_zip-2024-03-02.tar.gz (159228453 bytes, which is bigger than uncompressed enwiki8).

wiz21c24 hours ago

while impressive, it's a very specific case of compression: english text. It may be in use in many places but there are many more things to compress.

It'd be nice to have a comparison here: https://morotti.github.io/lzbench-web/?dataset=silesia/sao&m...

smusamashah1 day ago

Doesn't this fit the Hutter Prize conditions that is mentioned in other comment here https://news.ycombinator.com/item?id=46595109

WithinReason23 hours ago

It's too slow for that. The Hutter prize is CPU only so neural network solutions (which are the most interesting IMO) are effectively excluded. You need to generate 11 574 characters per second on the CPU only for decompression, and the compression time also counts and has to be below 24 hours in total.

cztomsik14 hours ago

IIRC all the cmix submissions are using NN (and were for long time)

WithinReason13 hours ago

well yes, small LSTMs I think, which is far from LLM territory

egl20201 day ago

When Jeff Dean gets stuck, he asks Bellard for help...

ok_dad1 day ago

Jeff Dean uses Fabrice Bellard as a rubber duck for debugging, and vice versa. Amazingly, neither says a thing for several minutes, staring into each other's eyes, and then they just start typing the solution to the problem on a single keyboard together.

jl623 hours ago

Shannon approaches the Bellard limit.

gmuslera1 day ago

Reminded me of pi filesystem (https://github.com/philipl/pifs), with enough digits of pi precalculated you might be able to do a decent compression program. The trick is in the amount of reasonable digits for that, if it’s smaller or bigger than that trained LLM.

GuB-421 day ago

I suspect that the length of the offset of your input data in pi is equal to the length of the input data itself, plus or minus a few bytes at most, regardless of the size of the input data.

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.

gmuslera20 hours ago

What if instead of the index of your full data, you store the index of smaller blocks? Would I need i.e. to use an 8kbytes or larger integer to store the offset all the possible 8k blocks?

It is meant to be a joke anyway.

sumtechguy19 hours ago

That would 'work' to a point. But my gut guess is it would end up with bigger data.

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 :)

noctune22 hours ago

Some patterns must happen to repeat, so I would assume the offset to be larger, no?

MrLeap1 day ago

You could express the offset with scientific notation, tetration, and other big math number things. You probably don't need the whole offset number all at once!

GuB-421 day ago

Actually, you do.

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.

miki12321121 hours ago

In that vein, pi is the true "illegal number"[1], as all the illegal content ever produced can eventually be found in the digits of pi.

I guess when God made the Universe, he filled it with CSAM, hate speech and unsolicited Viagra ads...

[1] https://en.wikipedia.org/wiki/Illegal_number

anthk19 hours ago

Pi is not bound to God neither the universe. It's just the ratio between a distance and adding up all the existing points around you at that distance.

SAI_Peregrinus18 hours ago

Which means it depends on the geometry of spacetime. If spacetime is quantized such that the Euclidean norm doesn't apply, then the value of pi will be different. See the "Extremal values of pi" paper[1].

[1]https://e.math.cornell.edu/people/Nikhil_Sahoo/files/pi_pape...

jlokier15 hours ago

Pi falls out of infinite sums, exponentials and differential equations all over the place. It's a universal constant, first discovered via geometry, but it's independent of geometry.

For 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''.

anthk17 hours ago

You can define Pi without universal geometry. And it's still a constant, nothing arbitrary, unlike

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     rot13
sltkr1 day ago

I'm going to be the nerd that points out that it has not been mathematically proven that pi contains every substring, so the pifs might not work even in theory (besides being utterly impractical, of course).

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.

dataflow1 day ago

> I'm going to be the nerd that points out that it has not been mathematically proven that pi contains every substring

Fascinating. Do you know if this has been proven about any interesting number (that wasn't explicitly constructed to make this true)?

cocoto13 hours ago

You can use the number 0.{concatenation of all 1 bit strings ordered}{concatenation of all 2 bits strings ordered} = 0.0100011011…

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.

kadoban1 day ago

https://en.wikipedia.org/wiki/Normal_number has some examples and a bunch of info. Most of them are pretty artificial, but the concatenation of the primes one is... at least interesting, not obvious (to me) from doing that that it'd be normal.

+1
_ache_23 hours ago
meindnoch18 hours ago

If we assume pi's digits to be uniformly random, then the expected offset for the first occurrence of a particular N-bit sequence is going to be ~2^N. (This can be proven using a Markov-chain argument. Also note: we're working in binary). So you've converted an N-bit value into an offset on the order of 2^N, which takes again N bits to represent.

eru1 day ago

> I'm going to be the nerd that points out that it has not been mathematically proven that pi contains every substring, so the pifs might not work even in theory (besides being utterly impractical, of course).

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.

netsharc1 day ago

You mentioning the concept of pi containing every substring makes me think of Borges' Library of Babel.

Ha, next: a compression algorithm that requires the user to first build an infinite library...

_ache_23 hours ago

Yep, the Library of Babel is a related topic. Wikipedia links it as "See also" on the page of "Normal numbers".

> 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...

https://github.com/philipl/pifs

charcircuit1 day ago

This only does 1 byte, so you only have to prove it contains the bits for 0-255.

meisel1 day ago

Looks like it beats everything in the large text compression benchmark for enwik8, but loses to several programs for enwik9. I wonder why that is.

AnotherGoodName1 day ago

It's actually not the best at enwik8 or 9.

The results at https://www.mattmahoney.net/dc/text.html explicitly add the size of the compressor itself to the result. Note the "enwik9+prog" column. That's what it's ranked on.

The reason to do this is that it's trivial to create a compressor that 'compresses' a file to 0 bytes. Just have an executable with a dictionary of enwik9 that writes that out given any input. So we always measure what is effectively the Kolmogorov complexity. The data+program as a whole that produces the result we want.

So those results add in the compressor size. The programs there generally have no dictionary built in or in the case of LLM based compressors, no pre-trained data. They effectively build the model as they process data. Not compressing much at all at the start and slowly compressing better and better as they go. This is why these programs do better and better with larger data sets. They start with 0 knowledge. After a GB or so they have very good knowledge of the corpus of human language.

This program here however is pre-trained and shipped with a model. It's 150MB in size! This means it has 150MB of extra starting knowledge over those models in that list. The top models in that list are the better compressors, they'll quickly out learn and overtake this compressor but they just don't have that headstart.

Of course measuring fairly this should be listed with that 150MB program size added to the results when doing a comparison.

srcreigh1 day ago

As an aside, I wonder how to account for the information content embedded in the hardware itself.

A Turing Machine compressor program would likely have more bytes than the amd64 binary. So how to evaluate KolmogorovComplexity(amd64)?

The laws of physics somehow need to be accounted for too, probably.

d_burfoot1 day ago

Kolmogorov Complexity is only defined up to a constant, which represents Turing machine translation length.

+1
notpushkin1 day ago
Dylan168071 day ago

The complexity of a simple turing machine is itty bitty, and you can bootstrap that into an x86 emulator in a matter of kilobytes, so when we're messing with 100MB files it's not a big factor.

rao-v1 day ago

If every version of your OS ships with a default local model, it maybe interesting to see compression conditioned on the standard LLM weights

oxag3n1 day ago

Compression and intelligence reminded me of the https://www.hutter1.net/prize

I've encountered it >10 years ago and it felt novel that compression is related to intelligence and even AGI.

eru1 day ago

Yes.

When you train your neural network to minimise cross-entropy that's literally the same as making it better as a building block in an arithmetic coding data compressor. See https://en.wikipedia.org/wiki/Arithmetic_coding

See also https://learnandburn.ai/p/an-elegant-equivalence-between-llm...

senderista1 day ago

Indeed, KL-divergence can be seen as the difference between the average number of bits required to arithmetically encode a sample from a given distribution, using symbol probabilities from both the original distribution and an approximating distribution.

https://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_diver...

wewewedxfgdf1 day ago

>> The ts_zip utility can compress (and hopefully decompress) text files

Hopefully :-)

hamandcheese1 day ago

Reading data is overrated. I highly recommend S4:

http://www.supersimplestorageservice.com/

fcantournet14 hours ago

It's particularly well suited for backups

MisterTea1 day ago

This is something I have been curious about in terms of how an LLM's achieves compression.

I would like to know what deviations are in the output as this almost feels like a game of telephone where each re-compression results in a loss of data which is then incorrectly reconstructed. Sort of like misremembering a story and as you tell it over time the details change slightly.

Scaevolus1 day ago

When LLMs predict the next token, they actually produce a distribution of the probability of each of the possible next tokens, and the sampler chooses one of them, and not necessarily the most likely one!

If instead you run LLM prediction and then encode the probability of the next token of the input text you want to encode (from the cumulative distribution, a number in [0, 1]) using arithmetic coding, you can run the same operation in reverse to achieve lossless compression.

The tricky part is ensuring that your LLM executes absolutely deterministically, because you need to make sure that the encoder and decoder have the same probability distribution map at each step.

AnotherGoodName1 day ago

Yes. The secret is in understanding arithmetic coding. https://en.wikipedia.org/wiki/Arithmetic_coding

Arithmetic coding takes a prediction of the next bit and writes out exactly as many bits as needed to correct that prediction. The amazing part is that you can write out fractional bits. Eg. You predict the next bit is '1' with 75% probability? If it is 1 you only need to write out 1/2 of a bit (correcting that 25% portion). If it's 0 you need to write out 2.4bits. It may seem strange to work with 1/2 a bit but it works! (essentially the other half of the bit represents other future correction required). You might have heard of huffman coding which can't deal with fractional bits, arithmetic coding is a generalization of huffman coding that can deal with this.

Arithmetic coding is mathematically perfect at what it does. You will not waste a single bit using this algorithm to encode data given a prediction of that data.

So the entirety of modern compression techniques don't deal with the encoding/decoding side at all. What they deal with is modelling the data so far and making the most accurate prediction possible on the next bit of data (next byte also works, but working 1 bit at a time is easier to comprehend when learning arithmetic coding).

Incidentally the encoders and decoders essentially work exactly the same. Given the data read or data decoded so far predict the next bit. This part is exactly the same either way. The decoder would read the compressed file for the correction and the encoder would read the input file and write out the correction. The important part is "predict the next bit". This is what separates all the different compressors.

This is also where those of us experienced in this area try to correct people on the wrong track. A compression algorithm is never about the encoding side but instead 100% always about the prediction of the data. Can you build a model that can accurately predict what the next data to come is? That's what you need to do to make a better file compressor. The entropy encoding part is a completely solved problem already, don't bother re-solving that.

billforsternz1 day ago

I don't understand. On average, for every 4 input bits we will get it right 3 times writing 0.5 bits each time and get it wrong once writing 2.4 bits once. So we write a total of 3 * 0.5 + 2.4 bits = 3.9 bits. The compressed output is 3.9/4 = 97.5% as big as the input. Not very compelling. What am I misunderstanding?

AnotherGoodName1 day ago

I back of the enveloped it wrong is what :(.

It's -log2(0.75) for getting a 75% chance right and -log2(0.25) for getting it wrong. I should have stated .4 bits and 2bits respectively not 0.5 and 2.4. Sorry! Good catch.

It's 3.2 vs 4bits. Now that may not seem huge but the probabilities tend to be at the more extreme ends if the predictor is any good. Once you start going towards the 99% range you get extreme efficiency.

billforsternz1 day ago

Thanks for the clarification.

ChadNauseam1 day ago

That's true for lossless compression. Is there a generalization for lossy compression?

AnotherGoodName1 day ago

Lossy and lossless are essentially fundamentally the same actually. All state of the art lossy compressors use arithmetic coding as an example and they still do prediction. Eg. your favourite video codecs predict not only the next bit in the 2D frame, but also the next bit when modelling past frames (becomes a 3D cube of data at that point) and they also do things like motion prediction of individual objects in frame to help make a better prediction. They all use arithmetic encoders to encode the data.

Where the lossy part comes in is the point at which humans notice/don't notice data being thrown away. Got a bit that was waaay out of line in the prediction and going to cost you 10bits to correct? Perhaps to humans it doesn't matter? Can we throw it away? This throwing away of data is often done before the prediction+compression stage (eg. maybe quantizing the color space to 8bits from 32bits?) but from there on it's the same thing.

eru1 day ago

To simplify: lossy compression = lossless compression + a perception model that can tell you what aspect of the data you can safely toss away without anyone noticing.

+1
ChadNauseam1 day ago
shawnz1 day ago

Another fun application of combining LLMs with arithmetic coding is steganography. Here's a project I worked on a while back which effectively uses the opposite technique of what's being done here, to construct a steganographic transformation: https://github.com/shawnz/textcoder

akoboldfrying1 day ago

Cool! It creates very plausible encodings.

> The Llama tokenizer used in this project sometimes permits multiple possible tokenizations for a given string.

Not having tokens be a prefix code is thoroughly unfortunate. Do the Llama team consider it a bug? I don't see how to rectify the situation without a full retrain, sadly.

shawnz1 day ago

I can't imagine they consider it a bug, it is a common and beneficial property of essentially every LLM today. You want to be able to represent common words with single tokens for efficiency, but at the same time you still need to be able to represent prefixes of those words in the cases where they occur separately

akoboldfrying1 day ago

I find this surprising, but I suppose it must be more efficient overall.

Presumably parsing text into tokens is done in some deterministic way. If it is done by greedily taking the longest-matching prefix that is a token, then when generating text it should be possible to "enrich" tokens that are prefixes of other tokens with additional constraints to force a unique parse: E.g., if "e" is a token but "en" is too, then after generating "e" you must never generate a token that begins with "n". A text generated this way can be deterministically parsed by the greedy parser.

Alternatively, it would suffice to restrict to a subset of tokens that are a prefix code. This would be simpler, but with lower coding efficiency.

+1
shawnz1 day ago
bonzini1 day ago

I think it's plausible that different languages would prefer different tokenizations. For example in Spanish the plural of carro is carros, in Italian it's carro. Maybe the LLM would prefer carr+o in Italian and a single token in Spanish.

akoboldfrying20 hours ago

Certainly! What surprised me was that apparently LLMs are deliberately designed to enable multiple ways of encoding the same string as tokens. I just assumed this would lead to inefficiency, since I assumed that it would cause training to not know whether it should favour outputting, say, se|same or ses|ame after "open", and thus throw some weight on each. But provided there's a deterministic rule, like "always choose the longest matching token", this uncertainty goes away.

+1
bonzini15 hours ago
SnowProblem1 day ago

I love this because it gets to the heart of information theory. Shannon's foundational insight was that information is surprise. A random sequence is incompressible by definition. But what counts as surprise depends on context, and for text, we know a large amount of it is predictable slop. I suspect there's a lot of room to go along this style of compression. For example, maybe you could store an upfront summary that makes prediction more accurate. Or perhaps you could encode larger sequences or some kind of hierarchical encoding. But this is great.

bambax1 day ago

Yes! information is surprise, and that's why a measure of intelligence is the ability to predict.

bob10291 day ago

PPMd is the most exotic compressor I've actually used in production. The first time I saw it in action I thought it was lossy or something was broken. I had never seen structured text compress that well.

amelius19 hours ago

I didn't think about it much but I'm surprised it gives only a 50% reduction wrt normal compression.

rurban1 day ago

So did beat his own leading program from 2019, nncp, finally.

dmitrygr1 day ago

"compressed size" does not seem to include the size of the model and the code to run it. According to the rules of Large Text Compression Benchmark, total size of those must be counted, otherwise a 0-byte "compressed" file with a decompressor containing the plaintext would win.

underdeserver1 day ago

Technically correct, but a better benchmark would be a known compressor with an unknown set of inputs (that come from a real-world population, e.g. coherent English text).

eru1 day ago

Yes, definitely. Alas, it's just harder to run these kinds of challenges completely fairly and self-administered, than the ones where you have a fixed texts as the challenge and add the binary size of the decompressor.

paufernandez1 day ago

Yeah, but the xz algorithm is also not counted in the bytes... Here the "program" is the LLM, much like your brain remembers things by coding them compressed and then reconstructs them. It is a different type of compression: compression by "understanding", which requires the whole corpus of possible inputs in some representation. The comparison is not fair to classical algorithms yet that's how you can compress a lot more (given a particular language): by having a model of it.

wrs1 day ago

“Compressors are ranked by the compressed size of enwik9 (10^9 bytes) plus the size of a zip archive containing the decompresser.” [0]

[0] https://www.mattmahoney.net/dc/text.html

FartyMcFarter1 day ago

True for competitions, but if your compression algorithm is general purpose then this matters less (within reason - no one wants to lug around a 1TB compression program).

nailer15 hours ago

Silly queston: what does the 'ts' stand for?

nilstycho13 hours ago

I believe it's "TextSynth".

https://bellard.org/ts_server/

benatkin1 day ago

I propose the name tokables for the compressed data produced by this. A play on tokens and how wild it is.

fancyswimtime1 day ago

please pass the tokables to the left hand side

voidUpdate22 hours ago

> The ts_zip utility can compress (and hopefully decompress)

Ah yes, write-only memory

sylware17 hours ago

Maybe one day, for some data to compress, "AGI" with a semantic understanding of this data, will be able to write some this-very-data-to-compress specific code predictor and to generate the related compressed data stream (lossless and why not lossy).

:P

jokoon1 day ago

so barely 2 or 3 times better than xz

not really worth it

publicdebates1 day ago

Bellard finally working with his true colleague.