Lookup tables with precalculated things for the win!
In fact I don’t think we would need processors anymore if we were centrally storing all of the operations ever done in our processors.
Now fast retrieval is another problem for another thread.
crmd · 2h ago
Reminds me of when I started working on storage systems as a young man and once suggested pre-computing every 4KB block once and just using pointers to the correct block as data is written, until someone pointed out that the number of unique 4KB blocks (2^32768) far exceeds the number of atoms in the universe.
jodrellblank · 16m ago
Reminds me of when I imagined brute-forcing every possible small picture as simply 256 shades of gray for each pixel x (640 x 480 = 307200 pixels) = 78 million possible pictures.
Actually I don't have any intuition for why that's wrong, except that if we catenate the rows into one long row then the picture can be considered as a number 307200 digits long in base 256, and then I see that it could represent 256^307200 possible different values. Which is a lot: https://www.wolframalpha.com/input?i=256%5E307200
manwe150 · 17m ago
It seems like you weren’t really that far off from implementing it, you just need a 4 KB pointer to point to the right block. And in fact, that is what all storage systems do!
makmanalp · 1h ago
In some contexts, dictionary encoding (which is what you're suggesting, approximately) can actually work great. For example common values or null values (which is a common type of common value). It's just less efficient to try to do it with /every/ block. You have to make it "worth it", which is a factor of the frequency of occurrence of the value. Shorter values give you a worse compression ratio on one hand, but on the other hand it's often likelier that you'll find it in the data so it makes up for it, to a point.
There are other similar lightweight encoding schemes like RLE and delta and frame of reference encoding which all are good for different data distributions.
ww520 · 1h ago
The idea is not too far off. You could compute a hash on an existing data block. Store the hash and data block mapping. Now you can use the hash in anywhere that data block resides, i.e. any duplicate data blocks can use the same hash. That's how storage deduplication works in the nutshell.
valenterry · 1h ago
Except that there are collisions...
datameta · 1h ago
This might be completely naive but can a reversible time component be incorporated into distinguishing two hash calculations? Meaning when unpacked/extrapolated it is a unique signifier but when decomposed it folds back into the standard calculation - is this feasible?
mncharity · 1h ago
> if we were centrally storing all of the operations
Community-scale caching? That's basically what pre-compiled software distributions are. And one idea for addressing the programming language design balk "that would be a nice feature, but it's not known how to compile it efficiently, so you can't have it", is highly-parallel cloud compilation, paired with a community-scale compiler cache. You might not mind if something takes say a day to resolve, if the community only needs it run once per release.
chowells · 4h ago
Oh, that's not a problem. Just cache the retrieval lookups too.
michaelhoney · 3h ago
it's pointers all the way down
drob518 · 1h ago
Just add one more level of indirection, I always say.
EGreg · 1h ago
But seriously… the solution is often to cache / shard to a halfway point — the LLM model weights for instance — and then store that to give you a nice approximation of the real problem space! That’s basically what many AI algorithms do, including MCTS and LLMs etc.
EGreg · 1h ago
You’re not wrong
Using an LLM and caching eg FAQs can save a lot of token credits
AI is basically solving a search problem and the models are just approximations of the data - like linear regression or fourier transforms.
The training is basically your precalculation. The key is that it precalculates a model with billions of parameters, not overfitting with an exact random set of answers hehe
LPisGood · 6h ago
I think it is very intuitive that more space beats the pants off of more time.
In time O(n) you can use O(n) cells on a tape, but there are O(2^n) possible configurations of symbols on a tape of length n (for an alphabet with 2 symbols), so you can do so much more with n space than with n time.
hn_acc1 · 5h ago
My intuition: the value of a cell can represent the result of multiple (many) time units used to compute something. If you cannot store enough intermediate results, you may end up needing to recalculate the same / similar results over and over - at least in some algorithms. So one cell can represent the results of hundreds of time units, and being able to store / load that one value to re-use it later can then replace those same hundreds of time units. In effect, space can be used for "time compression" (like a compressed file) when the time is used to compute similar values multiple times.
If intermediate results are entirely uncorrelated, with no overlap in the work at all, that would not hold - space will not help you. Edit: This kind of problem is very rare. Think of a cache with 0 percent hit rate - almost never happens.
And you can't really do it the other way around (at least not in current computing terms / concepts): you cannot use a single unit of time as a standin / proxy for hundreds of cells, since we don't quite have infinitely-broad SIMD architectures.
slt2021 · 5h ago
I think this theorem applies well for modern LLMs: large language model with pre-computed weights can be used to compute very complex algorithms that approximate human knowledge, that otherwise were impossible or would have required many orders more compute to calculate
frollogaston · 4h ago
Also, the O(1) random memory access assumption makes it easy to take memory for granted. Really it's something like O(n^(1/3)) when you're scaling the computer to the size of the problem, and you can see this in practice in datacenters.
I forget the name of the O(1) access model. Not UMA, something else.
cperciva · 3h ago
O(n^(1/2)) really, since data centers are 2 dimensional, not 3 dimensional.
(Quite aside from the practical "we build on the surface of the earth" consideration, heat dissipation considerations limit you to a 2 dimensional circuit in 3-space.)
mpoteat · 2h ago
More fundamentally O(n^(1/2)) due to the holographic principle which states that the maximal amount of information encodable in a given region of space scales wrt its surface area, rather than its volume.
(Even more aside to your practical heat dissipation constraint)
frollogaston · 2h ago
If you have rows of racks of machines, isn't that 3 dimensions? A machine can be on top of, behind, or next to another that it's directly connected to. And the components inside have their own non-uniform memory access.
Or if you're saying heat dissipation scales with surface area and is 2D, I don't know. Would think that water cooling makes it more about volume, but I'm not an expert on that.
manwe150 · 11m ago
That example would be two dimensions still in the limit computation, since you can keep building outwards (add buildings) but not scale upwards (add floors)
jvanderbot · 2h ago
Spatial position has nothing (ok only a little) to do with topology of connections.
LegionMammal978 · 3h ago
On the other hand, actual computers can work in parallel when you scale the hardware, something that the TM formulation doesn't cover. It can be interesting which algorithms work well with lots of computing power subject to data locality. (Brains being the classic example of this.)
LPisGood · 2h ago
Multitape TMs are pretty well studied
thatguysaguy · 6h ago
Intuitive yes, but since P != PSPACE is still unproven it's clearly hard to demonstrate.
LPisGood · 5h ago
I think that since many people find it intuitive that P != NP, and PSPACE sits way on top of polynomial hierarchy that it is intuitive even if it’s unproven.
porphyra · 4h ago
There's not even a proof that P != EXPTIME haha
EDIT: I am a dumbass and misremembered.
LPisGood · 3h ago
I thought there was some simple proof of this, but all I can think of is time hierarchy theorem.
doc_manhat · 4h ago
I think there is right? It's been a long time but I seem to remember it following from the time hierarchy theorem
dragontamer · 5h ago
The article is about a new proof wherein P == PSPACE.
Something we all intuitively expected but someone finally figured out an obscure way to prove it.
--------
This is a really roundabout article that takes a meandering path to a total bombshell in the field of complexity theory. Sorry for spoiling but uhhh, you'd expect an article about P == PSPACE would get to the point faster....
LPisGood · 5h ago
This article is not about a proof that P = PSPACE. That would be way bigger news since it also directly implies P = NP.
qbane · 5h ago
But you also spend time on updating cells, so it is not that intuitive.
LPisGood · 5h ago
I’m not sure what you mean here. If you’re in the realm of “more space” than you’re not thinking of the time it takes.
More precisely, I think it is intuitive that the class of problems that can be solved in any time given O(n) space is far larger than the class of problems that can be solved in any space given O(n) time.
Almondsetat · 5h ago
If your program runs in O(n) time, it cannot use more than O(n) memory (upper bound on memory usage.
If your program uses O(n) memory, it must run at least in O(n) time (lower bound on time).
No comments yet
delusional · 5h ago
This is obviously demonstrably true. A Turing running in O(n) time must halt. The one in O(n) space is free not to.
thaumasiotes · 4h ago
Almondsetat's proof seems more obvious. Given O(n) time, you can only use O(n) space, so you're comparing "O(n) space, any amount of time" with "O(n) space, O(n) time", and it turns out you get more resources the first way.
I am confused. If a single-tape turing machine receives a digit N in binary, and is supposed to write N ones on the tape, on the right side of the digit N, it performs N steps.
If you expect N ones at the output, how can this machine be simulated in the space smaller than N?
This machine must decrement the digit N at the beginning of the tape, and move to the end of the tape to write "1", so it runs in time O(N^2), not O(N)? (as it takes N "trips" to the end of the tape, and each "trip" takes 1, 2, 3 .. N steps)
Since turing machines can not jump to any place on a tape in constant time (like computers can), does it have any impact on real computers?
cperciva · 3h ago
Multitape Turing machines are far more powerful (in terms of how fast they can run, not computability) than single-tape machines.
But to answer your question: "space" here refers to working space, excluding the input and output.
willmarquis · 2h ago
This is just a reminder that memory isn’t just a constraint, it’s a resource.
michaelsbradley · 43m ago
What resources are available (or not) and in what quantities are the most basic constraints for solving a problem/s with a computer.
https://arxiv.org/abs/2502.17779
In fact I don’t think we would need processors anymore if we were centrally storing all of the operations ever done in our processors.
Now fast retrieval is another problem for another thread.
Actually I don't have any intuition for why that's wrong, except that if we catenate the rows into one long row then the picture can be considered as a number 307200 digits long in base 256, and then I see that it could represent 256^307200 possible different values. Which is a lot: https://www.wolframalpha.com/input?i=256%5E307200
There are other similar lightweight encoding schemes like RLE and delta and frame of reference encoding which all are good for different data distributions.
Community-scale caching? That's basically what pre-compiled software distributions are. And one idea for addressing the programming language design balk "that would be a nice feature, but it's not known how to compile it efficiently, so you can't have it", is highly-parallel cloud compilation, paired with a community-scale compiler cache. You might not mind if something takes say a day to resolve, if the community only needs it run once per release.
Using an LLM and caching eg FAQs can save a lot of token credits
AI is basically solving a search problem and the models are just approximations of the data - like linear regression or fourier transforms.
The training is basically your precalculation. The key is that it precalculates a model with billions of parameters, not overfitting with an exact random set of answers hehe
In time O(n) you can use O(n) cells on a tape, but there are O(2^n) possible configurations of symbols on a tape of length n (for an alphabet with 2 symbols), so you can do so much more with n space than with n time.
If intermediate results are entirely uncorrelated, with no overlap in the work at all, that would not hold - space will not help you. Edit: This kind of problem is very rare. Think of a cache with 0 percent hit rate - almost never happens.
And you can't really do it the other way around (at least not in current computing terms / concepts): you cannot use a single unit of time as a standin / proxy for hundreds of cells, since we don't quite have infinitely-broad SIMD architectures.
I forget the name of the O(1) access model. Not UMA, something else.
(Quite aside from the practical "we build on the surface of the earth" consideration, heat dissipation considerations limit you to a 2 dimensional circuit in 3-space.)
(Even more aside to your practical heat dissipation constraint)
Or if you're saying heat dissipation scales with surface area and is 2D, I don't know. Would think that water cooling makes it more about volume, but I'm not an expert on that.
EDIT: I am a dumbass and misremembered.
Something we all intuitively expected but someone finally figured out an obscure way to prove it.
--------
This is a really roundabout article that takes a meandering path to a total bombshell in the field of complexity theory. Sorry for spoiling but uhhh, you'd expect an article about P == PSPACE would get to the point faster....
More precisely, I think it is intuitive that the class of problems that can be solved in any time given O(n) space is far larger than the class of problems that can be solved in any space given O(n) time.
If your program uses O(n) memory, it must run at least in O(n) time (lower bound on time).
No comments yet
And paper: https://people.csail.mit.edu/rrw/time-vs-space.pdf
If you expect N ones at the output, how can this machine be simulated in the space smaller than N?
This machine must decrement the digit N at the beginning of the tape, and move to the end of the tape to write "1", so it runs in time O(N^2), not O(N)? (as it takes N "trips" to the end of the tape, and each "trip" takes 1, 2, 3 .. N steps)
Since turing machines can not jump to any place on a tape in constant time (like computers can), does it have any impact on real computers?
But to answer your question: "space" here refers to working space, excluding the input and output.