7 ms·
New proof dramatically compresses space needed for computation
- Gehinnn 1y agoThe impossible chess board problem must have something to do with the idea of solving tree eval with little memory (https://youtu.be/wTJI_WuZSwE?si=lgTc65RhXQesKchR https://youtu.be/wTJI_WuZSwE?si=lgTc65RhXQesKchR)! When the chess board is random, it feels impossible to add additional information. However, by choosing which cell to flip and knowing that you followed a certain operation, you can encode the position information in the random data, without needing more cells.
- baruchel 1y agoWithout paywall: https://www.removepaywall.com/search?url=https://www.scientificamerican.com/article/new-proof-dramatically-compresses-space-needed-for-computation/ https://www.removepaywall.com/search?url=https://www.scienti...
- trueismywork 1y agohttps://arxiv.org/abs/2502.17779 https://arxiv.org/abs/2502.17779 Link to paper
- zerof1l 1y agoHere's the gist: For nearly 50 years theorists believed that if solving a problem takes t steps, it should also need roughly t bits of memory: 100 steps - 100bits. To be exact t/log(t). Ryan Williams found that any problem solvable in time t needs only about sqrt(t) bits of memory: a 100-step computation could be compressed and solved with something on the order of 10 bits.
- zombot 1y ago> log(t) log to what basis? 2 or e or 10 or... Why do programmers have to be so sloppy?
- 12345ieee 1y agoIt doesn't matter in O() notation.
- Tarq0n 1y agoThis is very common. Log without further specification can be assumed to be the natural log (log e).
- griffzhowl 1y agoIn information theory it usually means log base 2
- eviks 1y agono, that's what ln is for
- thaumasiotes 1y agoWell, you're right that that's what "ln" is for. But more specifically "ln" is for indicating the natural log on calculators that already have another button labeled "log". Tarq0n is correct that "log" without further specification can be assumed to be the natural log.
- holowoodman 1y agoNo. Usually log without further specification is base10. Except in mathematics and physics, where it usually is base e. Except sometimes in computer science, where it can be base 2. But there are more of those weirdnesses: "ld" can be "log decimal", so base 10; or "logarithmus dualis", so base 2. Base 2 is also sometimes written as "lb" (log binary). You really need to know the conventions of the field/subfield to know which is which.
- eru 1y agoLog without a basis (at least to me) usually indicates that the basis doesn't matter in this context (or is obvious). I usually see either base e or base 2 as the default. In which applications do people use both logarithms and base 10 as the default? I mean, especially these day. In the olden days of the slide ruler base 10 was probably more common.
- heavenlyblue 1y agoWhy does it need the same bits of memory as steps?
- mjburgess 1y agoThe basic intuition of the two extremes are: * High space requirement: we need a bucket per step to track the state * Low space requirement: we only need a single bucket, if we can mix/unmix the various components of the state High-space requirement algorithms will be those with a large amount of "unmixable" complex state. Low-space reqs will be those with either a small state, or an easily mixed/unmixed state. Example of mixing: suppose we need an odd number and a parity (+,-) -- then we can store odd/even numbers: taking even to means (-1 * (number-1)) and odd means odd. So 7 = +7, 8 = -7 I don't know if this example is really that good, but I believe the very basics of the intution is correct -- its to do with how we can trade "interpretation" of data (computation) for the amount of data in such a way that data can kinda be mixed/unmixed in the interpretation
- TrueDuality 1y agoBits are a bit misleading here, it would be more accurate to say "units of computation". If you're problem space operates on 32-bit integers this would be 32bits * number of steps, these papers solve for individual bits as the smallest individual unit of computation we can commonly reason about.
- mort96 1y agoWait but if "bits of memory" here was supposed to be "units of computation", that means that: * The old assumption was that if you require t steps to complete, you need t/log(t) units of computation * This new proof shows that if you require t steps to complete, you need sqrt(t) units of computation Surely this doesn't make sense? Using any definition of "unit of computation" I would intuitively assume, computing t steps requires something proportionalt to t units of computation...
- nyrikki 1y agoThere is a bit of nuances here that is difficult to explain fully but note from the paper. > Our simulation reduces the problem of simulating time-bounded multitape Turing machines to a series of implicitly-defined Tree Evaluation instances with nice parameters, leveraging the remarkable space-efficient algorithm for Tree Evaluation recently found by Cook and Mertz [STOC 2024]. The "time-bounded multitape Turing machines" with bounded fan-in means that that particular abstract model has access to the bits of those tapes current head position. Mapping the quirks of the various abstract models can be tricky, but remember having access to the symbol currently under the tape head is 'free' in TMs. It is a useful abstraction but doesn't directly map to physically realizable machine. It is still an interesting result for trees in physically realizable machines.
- taneq 1y agoHuh? What did any of those theorists think about the phrase "time/space tradeoff"?
- mhuffman 1y agoThey were likely the ones that invented the phrase.
- taneq 1y agoI'm not convinced of that if they thought that s=O(t/log(t)) where s=space (bytes or similar) and t=time (clocks, steps, etc.) I dunno about theoretical limits but in practice, there's at least 1-2 orders of magnitude to be exchanged between space and processing time.
- eru 1y agoThat's in practice for practical problems. But they were interested in the worst cases.
- mort96 1y agoReally? But the whole idea behind the space/time trade-off is that in a bunch of cases, if you want to compute something in less memory, you need to solve it in more steps; or if you want to solve something in fewer steps, you need to use more memory. This seems to wildly contradict the idea that the amount of memory needed is roughly proportional to the number of steps.
- eru 1y agoThe amount of steps is an (over)estimate on the amount of memory needed. You can write some trivial programs that need this much memory. But most of them can be optimised to use less space (and still compute the same answer). The big question was: can you always optimise the space usage down from the ridiculous maximum and to what extent?
- cubefox 1y ago> Ryan Williams found that any problem solvable in time t needs only about sqrt(t) bits of memory At least or at most or on average?
- user070223 1y agoI've skimmed the result couple of weeks ago, and if I remember it states that there from all machines which takes t time to accomplish something there is one such that sqrt(t) memory is used. so it's at most but not on avg nor amorotized[not sure if et even make sense it the space where the problem lies] (you take the least memory hungry machine)
- cma 1y agoThere must be more constraints, what if the problem is copying a string? Or is it only for turing tape machines where it has to backtrack during the copy, increasing number of steps?
- jasperry 1y agoThat's a good example! So the sqrt(n) space in the result must refer to space beyond that needed to write down the output--basically a working memory scratchpad. In which case, a string-copying function could use just a constant amount of scratch space. I mean, for all I know, the result may have been proved in the model of decision problems where the output is always one bit. But I'd guess that it generalizes just fine to the case where you have to write longer outputs. However, your question does make me wonder if the result still applies in the RAM model, where you can move to any spot on the tape in constant time. My theory knowledge is getting rusty...
- jasperry 1y agoUpdate after youtube: It was already known that a time-t function can be computed in sqrt(t) space on a single-tape Turing machine. It's exactly like @cma said: a Turing machine wastes a lot of time moving the head back and forth. Williams' result shows sqrt(t) space for multi-tape Turing machines. Those are much more time-efficient in regard to head movement, so until Williams' proof it was not believed to be possible to to make any algorithm use only sqrt(t) space.
- m3kw9 1y agoDoesn’t make practical sense why they would even assign a number to problems which could have unknown dependencies. Unless you are talking about bounded math issues
- andsoitis 1y ago> Ryan Williams found that any problem solvable in time t needs only about sqrt(t) bits of memory: a 100-step computation could be compressed and solved with something on the order of 10 bits. Does it require one to first have solved the problem in uncompressed space?
- jasperry 1y agoAs I understand it, this result is about algorithms that solve any instance of a given computational problem (like finding the shortest paths through a graph.) The specific problem instance (the graph) is the input to the algorithm. The result shows how to take an algorithm and construct another algorithm that solves the same class of problems in less space, by transforming how the algorithm reads and writes memory. Now you have an algorithm that still solves any instance of that problem but uses only sqrt(t) space. So, to answer your question more directly: no, you don't have to have solved a specific instance of the problem, but you have to already have an algorithm that solves it in general.
- svat 1y agoTo be clear, this was/is a statement about the worst case, not every problem. So it may be clearer to rephrase your sentence as: For nearly 50 years theorists believed that there exist problems taking t steps that also need roughly t/log(t) bits of memory. (The Quanta magazine article https://www.quantamagazine.org/for-algorithms-a-little-memory-outweighs-a-lot-of-time-20250521/ https://www.quantamagazine.org/for-algorithms-a-little-memor... gives some context: for a long time, the best result was t/log(t) as proved in "On Time Versus Space" https://dl.acm.org/doi/10.1145/322003.322015 https://dl.acm.org/doi/10.1145/322003.322015 by Hopcroft, Paul and Valiant in 1975.)
- makira 1y agoSee previous discussion: https://news.ycombinator.com/item?id=44055347 https://news.ycombinator.com/item?id=44055347
- cyrillite 1y agoOne of my favourite sci comms YouTubers explained this in great detail https://youtu.be/8JuWdXrCmWg https://youtu.be/8JuWdXrCmWg Highly recommend
- teekert 1y agoYeah, where Hossenfelder is getting more and more dramatic (although I can still appreciate it) she has a refreshingly calm and intelligent tone. Highly recommended indeed.
- eru 1y agoSabine Hossenfelder also has very 'interesting' ideas about determinism.
- teekert 1y agoWhich I share. It’s haunted me for a long time, but I’ve accepted it. Much like Sabine. We can’t predict the future but we do not have free will as most people think we have, imho. Many of those separated brain cases seem to confirm this stance.
- hnuser123456 1y agoSure, but if someone finds themselves feeling incredibly defeated by the thought, then how can we call it productive philosophy? I went too far down this rabbit hole about 8 years ago, and built up a strong mindset of all the things I wanted to be that I couldn't because I wasn't born in the right circumstances. Much better to feel like I can absolutely be those things at least in spirit, and maybe talk to other people about it and find people who are willing to see the unexpected parts of me. Yes, we have enough accurate theories now that we can predict parts of the future with incredible accuracy in ways that weren't possible 100+ years ago, but we don't have a bulletproof theory of everything, much less a bulletproof theory of everything about humans. Championing superdeterminism is like being the smart alec who says they can predict anything if you give them enough initial context. Sure, now go make risky investments that will only go up. The Heisenberg uncertainty principle itself shows that it is not worth fretting too much about superdeterminism. We will never be able to replace every theory that uses probabilities with ones that don't.
- kristianp 1y agoThis seems very theoretical, just a lower bound on space required, without talking about what is being computed. Does it have any import on real algorithms?
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- wiz21c 1y agoLower bound tells you how much it's worth to improve the SOTA. It gives you a hint that you can do better. So it's more like polar star. Maybe not directly practical, but it will lead tons of people in the right direction.
- Gravityloss 1y agoThanks. I think the title is quite misleading then? I would have expected better from Scientific American.
- kadoban 1y ago> Does it have any import on real algorithms? As-in algorithms you'll care about if you're a programmer and not overly interested in theory? Not really, no.
- CommenterPerson 1y agoIt does have an impact. It wont give you stackoverflow like code to copy-paste though. Analogy is thermodynamics says how efficient a heat engine _could_ be. If your engine is way below that you know there how much of an improvement there _could_ be, that it's not an impossible problem. That will get people to build better stuff.
- bmacho 1y agoUpper bound on the space required. Anything, that is computable. > Does it have any import on real algorithms? It depends on the constants, but if the constants are good, you can have VMs for example that make memory usage smaller.
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- mikewarot 1y agoI get that this is an interesting theoretical proof. I'm more interested in the inverse, trading memory for time, to make things faster, even if they take more space. It seems to me the halting problem is almost a proof the inverse of this is impossible. Memoization is likely the best you can do.
- LegionMammal978 1y agoThe author of the paper also notes how the result gives an upper bound for how well we can 'trade space for time' in the worst case: > In this paper, we make another step towards separating P from PSPACE, by showing that there are problems solvable in O(n) space that cannot be solved in n^2/log^c n time for some constant c > 0. [0] Of course, this is all specifically in the context of multitape Turing machines, which have very different complexity characteristics than the RAM machines we're more used to. [0] https://arxiv.org/abs/2502.17779 https://arxiv.org/abs/2502.17779
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- burnt-resistor 1y agoDynamic programming and divide-and-conquer with multiple cores on a diagonalized problem space of independent problems (CPU and memory hierarchy) with (hopefully) a final reduce step at the end. Real-world problems are seldom so neat, hence the investment in Infiniband gear.
- JohnKemeny 1y agoRelated: For algorithms, a little memory outweighs a lot of time. 343 points, 139 comments, 39 days ago. https://news.ycombinator.com/item?id=44055347 https://news.ycombinator.com/item?id=44055347 You need much less memory than time. 126 points, 11 comments, 22 days ago. https://news.ycombinator.com/item?id=44212855 https://news.ycombinator.com/item?id=44212855
- ALLTaken 1y agoI am trying to understand how the paper OP posted can be translated into anything that improves current algorithms. Can actually any current algorithms improved by this?? Or is that only the case if the hardware itself is improved? I am baffled
- JohnKemeny 1y agoThis is blue skies research and is not expected to have any immediate practical impact whatsoever.
- utopcell 1y agoThe result doesn't actually apply to _all_ algorithms, only oblivious ones. Oblivious algorithms are ones where step i does not depend on decisions made in steps j < i. Almost all useful algorithms are non-oblivious, with some notable exceptions [1]. This work is a step forward towards proving the result for the general case, with little practical usage (and that's perfectly okay). [1] https://en.wikipedia.org/wiki/Sorting_network https://en.wikipedia.org/wiki/Sorting_network
- bluenose69 1y agoHere's a quote from the SciAm article: "Technically, that equation was t/log(t), but for the numbers involved log(t) is typically negligibly small." Huh?
- burnt-resistor 1y agoMaybe I'm missing context, but that sounds like O(n) or Ω(n).
- fwip 1y agot/log(t) is 'closer to' t than it is to sqrt(t) as t heads toward infinity. e.g: 4/log2(4) = 4/2 = 2 sqrt(4) = 2 2^32/log2(2^32) = 2^32/32 = 2^27 sqrt(2^32) = 2^16
- tgv 1y agoIn case someone doesn't like the proof by example, here's a hint: sqrt(t) = t / sqrt(t).
- asimpletune 1y agoI think this means that while Log grows to infinity, it does that so slowly that it can often be treated as if it were a coefficient. Coefficients are ignored in big O notation, hence the negligibly small comment.
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- gwbas1c 1y agoOne thing that's surprised me throughout my career is just how inefficient most of the code that I've inherited is. I suspect we could do a lot more with the hardware we have if we simply became better at programming. (Perhaps more AI coaching could help?)
- HideousKojima 1y agoAI is trained on the same shitty code you're inheriting, so probably not.
- humanfromearth9 1y agoIt's also trained on all best practices and algorithms that you don't know exist, so it is able to do better - provided you know to ask and how to ask/what to ask.
- HideousKojima 1y agoIt's not simply a matter of knowing what/how to ask. LLMs are essentially statistical regressions on crack. This is a gross oversimplification, but the point is that what they generate is based on statistical likelihoods, and if 90%+ of the code they were trained on was shit you're not going to get the good stuff very often. And if you need an AI to help you do it you won't even be able to recognize the good stuff when it does get generated.
- jiehong 1y agoI'm also surprised by how many developers just don't seem to care much. Yet, the general population isn't behaving like scouts most of the time, so I suppose it's only fair that developers also aren't. "Leave the world a better place than you found it" (s/world/code/g)
- Agingcoder 1y agoI’ve spent many years working on hpc problems, and most people simply don’t have , or don’t care about, or don’t know what to do with performance problems. Don’t have to because they usually have a full machine, and unless they saturate it, the problem doesn’t exist. Don’t care, because many people think it’s alright if things are slow, or inefficient ( the user can wait, or will live with 100ms instead of 1ms). Don’t know what to do - this one is interesting : many people don’t realize it’s actually possible to get the same result , but faster.
- krackers 1y agoRelated: https://news.ycombinator.com/item?id=44212855 https://news.ycombinator.com/item?id=44212855 and https://news.ycombinator.com/item?id=44055347 https://news.ycombinator.com/item?id=44055347
- gweinberg 1y agoI don't understand what they are saying at all. If I have a computation that requires going through a loop n times, why should the memory requirements increase with n at all?
- jasperry 1y agoSure, there are plenty of computations where you don't need much space no matter how many times you loop--like if you're just updating a single sum. But there are other problems where each time through the loop you might need to add an element to a big array or matrix, and then you go back and do more computation on the saved data elements to get the answer. Before this result, there were believed to be at least some problems where you need (almost) as much space as time, but Williams shows that, theoretically at least, that's never necessarily the case.
- svat 1y agoOf course in many cases (such as the one you mentioned), the amount of memory needed does not grow with N. But there are other programs/computation that seem to require more memory. For example, consider the following (Python) program: def intpow(n, r): return int(n**r) N = 10000 a = [0] * N a[1] = 1 for m in range(2, N): a[m] = a[m-1] + a[intpow(m, 0.6)] + a[intpow(m, 0.3)] + a[intpow(m, 0.1)] print(a[N-1]) It populates an array of size N by, for each index m, accessing array elements at smaller values m-1, m^0.6, m^0.3 and m^0.1 (just arbitrary values I made up). So this is a program that runs in time about N, and as it can depend on array elements that go very far back, it may not be immediately obvious how to make it run with space significantly less than N^0.9 (say), let alone √N. But the result shows that it can be done (disclaimer: I haven't actually verified the details of how the proof carries over to this program, but it looks like it should be possible), and for any program no matter how weird.
- snickerbockers 1y ago>(Technically, that equation was t/log(t), but for the numbers involved log(t) is typically negligibly small.) My dude, that is NOT how rational numbers work.
- utopcell 1y ago[ made this a top-level comment as I don't see it mentioned in other comments: ] The result doesn't actually apply to _all_ algorithms, only oblivious ones. Oblivious algorithms are ones where step i does not depend on decisions made in steps j < i. Almost all useful algorithms are non-oblivious, with some notable exceptions [1]. This work is a step forward towards proving the result for the general case, with little practical usage (and that's perfectly okay). [1] https://en.wikipedia.org/wiki/Sorting_network https://en.wikipedia.org/wiki/Sorting_network
- LegionMammal978 1y agoIt applies to any algorithm that can be implemented with a multitape Turing machine, which is still quite a few of them. The obliviousness requirement is only for random-access machines.
- aaron695 1y ago[dead]