Fabrice Bellard's TS Zip (2024)

(bellard.org)

192 points | by everlier 18 hours ago

18 comments

  • omoikane 15 hours 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).

    • wiz21c 6 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...

    • smusamashah 8 hours ago
      Doesn't this fit the Hutter Prize conditions that is mentioned in other comment here https://news.ycombinator.com/item?id=46595109
      • WithinReason 5 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.
  • gmuslera 15 hours 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.
    • miki123211 4 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

      • anthk 1 hour 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.
    • GuB-42 15 hours 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.

      • luke5441 2 hours ago
      • gmuslera 2 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.

        • sumtechguy 1 hour 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 :)

      • noctune 4 hours ago
        Some patterns must happen to repeat, so I would assume the offset to be larger, no?
      • MrLeap 13 hours 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-42 12 hours 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.

    • sltkr 14 hours 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.

      • meindnoch 29 minutes 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.
      • dataflow 10 hours 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)?

      • eru 13 hours 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.

      • netsharc 11 hours 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_ 5 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

      • charcircuit 14 hours ago
        This only does 1 byte, so you only have to prove it contains the bits for 0-255.
  • egl2020 16 hours ago
    When Jeff Dean gets stuck, he asks Bellard for help...
    • ok_dad 7 hours 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.
      • jl6 5 hours ago
        Shannon approaches the Bellard limit.
  • meisel 16 hours 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.
    • AnotherGoodName 15 hours 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.

      • srcreigh 15 hours 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_burfoot 15 hours ago
          Kolmogorov Complexity is only defined up to a constant, which represents Turing machine translation length.
          • notpushkin 10 hours ago
            I guess we need to guesstimate the length of a shortest Turing machine implementation of amd64 then?
        • Dylan16807 10 hours 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-v 12 hours 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
  • amelius 1 hour ago
    I didn't think about it much but I'm surprised it gives only a 50% reduction wrt normal compression.
  • oxag3n 16 hours 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.

  • wewewedxfgdf 17 hours ago
    >> The ts_zip utility can compress (and hopefully decompress) text files

    Hopefully :-)

  • rurban 16 hours ago
    So did beat his own leading program from 2019, nncp, finally.
  • MisterTea 17 hours 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.

    • Scaevolus 17 hours 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.

    • AnotherGoodName 15 hours 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.

      • billforsternz 12 hours 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?
        • AnotherGoodName 11 hours 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.

      • ChadNauseam 14 hours ago
        That's true for lossless compression. Is there a generalization for lossy compression?
        • AnotherGoodName 14 hours 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.

          • eru 13 hours 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.
          • ChadNauseam 13 hours ago
            Thanks! That's really enlightening. Maybe this can help me settle a question I've had. If I have 4k video and I want a smaller file size (but suppose I still have a 4k tv), I always felt like I should get better quality at that file size by compressing it further than by reducing the resolution. Rather than deciding for myself what data to throw away, why not let the compression algorithm do it?
            • AnotherGoodName 12 hours ago
              Lowering the resolution to match the typical TV resolutions is sensible but beyond that trusting the codec is always the better way. The codecs will adaptively compress portions of the video differently. The top left area that's in shadow for the next 30seconds? Maybe that areas effective resolution can be lowered? Etc. They can do things like this!

              If you change the resolution or color space of the entire file you do that without consideration to where the extra details might have been needed.

              So resolution should match typical output resolutions exactly and from there it's all on the codec.

    • themafia 13 hours ago
  • bob1029 12 hours 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.
  • shawnz 17 hours 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
    • akoboldfrying 13 hours 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.

      • shawnz 9 hours 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
        • akoboldfrying 9 hours 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.

          • shawnz 8 hours ago
            Regarding the first part: that's an interesting idea, although I worry it would bias the outputs in an unrealistic way. Then again, maybe it would only impact scenarios that would have otherwise been unparsable anyway?

            Regarding the second part: you'd effectively just be limiting yourself to single character tokens in that case which would drastically impact the LLM's output quality

            • akoboldfrying 6 hours ago
              The first approach would only affect outputs that would have been otherwise unparseable.

              The second approach works with any subset of tokens that form a prefix code -- you effectively set the probability of all tokens outside this subset to zero (and rescale the remaining probabilities if necessary). In practice you would want to choose a large subset, which means you almost certainly want to avoid choosing any single-character tokens, since they can't coexist with tokens beginning with that character. (Choosing a largest-possible such subset sounds like an interesting subproblem to me.)

              • shawnz 3 hours ago
                I don't think I see the vision here. If you want to maximize the number of tokens representable as a prefix code while still being able to output any sequence of characters, how could you possibly pick anything other than the one-character-long tokens?

                Are you saying you'd intentionally make some output sequences impossible on the basis they're not likely enough to be worth violating the prefix code for? Surely there's enough common short words like "a", "the", etc that that would be impractical?

                And even excluding the cases that are trivially impossible due to having short words as a prefix, surely even the longer words share prefixes commonly enough that you'd never get tokens longer than, say, two characters in the best case? Like, so many words start with "st" or "wh" or "re" or whatever, how could you possibly have a prefix code that captures all of them, or even the most common ones, without it being uselessly short?

                • akoboldfrying 2 hours ago
                  > Surely there's enough common short words like "a", "the", etc that that would be impractical?

                  Tokens don't have to correspond to words. The 2-character tokens "a " and " a" will cover all practical uses of the lowercase word "a". Yes, this does make some strings unrepresentable, such as the single-character string "a", but provided you have tokens "ab", "ba", "ac", "ca", etc., all other strings can be represented. In practice you won't have all such tokens, but this doesn't materially worsen the output provided the substrings that you cannot represent are all low-probability.

                  • shawnz 2 hours ago
                    Ah yeah, factoring in the whitespace might make this a bit more practical
      • bonzini 9 hours 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.
        • akoboldfrying 2 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.
  • SnowProblem 16 hours 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.
    • bambax 15 hours ago
      Yes! information is surprise, and that's why a measure of intelligence is the ability to predict.
  • dmitrygr 17 hours 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.
    • underdeserver 17 hours 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).
      • eru 13 hours 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.
    • FartyMcFarter 17 hours 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).
    • paufernandez 17 hours 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.
  • voidUpdate 5 hours ago
    > The ts_zip utility can compress (and hopefully decompress)

    Ah yes, write-only memory

  • benatkin 17 hours ago
    I propose the name tokables for the compressed data produced by this. A play on tokens and how wild it is.
    • fancyswimtime 16 hours ago
      please pass the tokables to the left hand side
  • jokoon 15 hours ago
    so barely 2 or 3 times better than xz

    not really worth it

  • publicdebates 17 hours ago
    Bellard finally working with his true colleague.