10 ms·

Computer memory prototype ditches 1s and 0s for denser data storage

failuser 3y ago
Vinyl records also don’t have 0’s and zeroes, it does not make them better than Blu-ray. MLC NAND have more than just 1’s and 0’s, they have 0..7 or even 0..15. Does anyone really want to process analog data?
- vlovich123 3y ago
  > Does anyone really want to process analog data? I imagine at some point in the abstractions it'll get a compat layer. Also, I could be wrong about MLC NAND but I thought it still stored 0/1 and just had more bits stored per cell through stacking.
  - whatevaa 3y ago
    No, multi-level NAND has extra voltage levels. Instead of a just regular off (0) and on (1) they have intermediate states too. For QLC, that's 16 levels, it's quite granular, and the margins of error are very small.
    - Retr0id 3y ago
      The margins for error are wider than one might immediately assume, since they combine it with error correction codes.
  - sp332 3y ago
    TLC NAND stores one of eight values in the cell. There aren't three separate values like 1, 0, 1, there is only one value, like "5".
    - AussieWog93 3y ago
      I mean, 101 is just 5 in binary. So each cell stores either three pieces of binary or one piece of octal. It's the same thing.
      - failuser 3y ago
        5 is not stored as 101, it’s stored as charge level number 5 out of 8 other valid charge levels.
      - koromak 3y ago
        Its absolutely not the same thing from a hardware perspective. Normal components always operate on bits. We see "5", but the computer sees (at least) 3 bits. It then sends its output as bits to the next one. A 5 must require 3 wires, or 3 cycles of the clock over a single wire. Building a computer where each wire can encode multiple bits of data at the exact same time fundamentally changes the way operations are done. The entire thing would need to be built in a different way, with different consequences. It might be a bad idea, or impossible, but it would be different.
- mcdonje 3y ago
  1. Vinyl ridges are much larger than DVD lines. Kind of a silly comparison considering the article claims the density of this scheme would exceed what we currently have. 2. It appears to not be analog. >An energy barrier is created at the points where the bridges meet the device contacts, and the height of this barrier can be controlled which changes the electrical resistance of the overall material. That in turn is what encodes the data. That seems to indicate they can have a base-n encoding where n is what they can practically achieve with the barrier/bridge layering scheme.
  - failuser 3y ago
    They claim it’s continuous. If continuous values itself were that great then the vinyl records were great for data storage. The problem with continuous values is noise, durability and precision. You can made vinyl disks with incredibly high precision, but preventing damage and reading back those values is not practical. That’s why even the first analog laser medium had discrete pits and lands instead of a continuous signal.
    - AnotherGoodName 3y ago
      It can be continuous but still have a limit to data due to noise. In vinyls case the more data you store the higher the error rate (at some point the differences in states is hidden in noise). The data rate from a noisy channel or storage can be derived from Shannon-hartley theorem. This is why the old analogue phones could only do 56k for example.
      - failuser 3y ago
        Yes, that’s why I hate their headline.
      - coldtea 3y ago
        >This is why the old analogue phones could only do 56k for example. And somehow, on analogue circuit switch phone call mode, still sounded better with less artifacts than most mobile phone calls!
- PhasmaFelis 3y ago
  > Does anyone really want to process analog data? If they can get it to encode binary data more densely than existing systems, with acceptable speed/reliability/cost? Yes, of course they would.
- kstrauser 3y ago
  My understanding is that modern hard drives all use probability and error detection/correction to convert a gooey analog signal into chunky discrete goodness.
- dragontamer 3y ago
  > Does anyone really want to process analog data? I mean... maybe we should? All radio waves, such as 802.11, are fundamentally analog data. And then they are converted back into digital with incredible QAM-encoding levels. If you think QLC 4-bits per Flash-cell is big, wait till you see 256-QAM (aka: 8-bits per 802.11 timeslice), or more. I believe I saw some spec for some fiber-optic cable that is 32768-QAM (aka: 15-bits per timeslice). -------- Then again, Flash (and other memory) are composed of individual cells. True-analog signals like radio-waves are truly analog, so maybe its not quite the same. But maybe we should think of things like tape or hard-drives as an analog signal. Or maybe not, who knows? I'm not actually a storage specialist so I don't really know. (Nor am I really in communications, lol).
  - foobiekr 3y ago
    256+ QAM is there but you will almost never see it in practice unless you’re next to the AP. Wifi7 defines up to 4096 so maybe the economics will drive improvements that didn’t land in 6E APs.
    - tguvot 3y ago
      docsis. up to 16384
    - Taniwha 3y ago
      You'll see it on almost every cableTV channel these days
      - foobiekr 3y ago
        I was referring to Wifi but yes
  - enkid 3y ago
    I'm pretty sure that Einstein proved that light (including radio) is not, in-fact, analog, but rather has discrete levels of energy in quantum steps.
- hn_throwaway_99 3y ago
  I think this is a faulty analogy. The data is still encoded (or, at least, interpreted) as digital data, except that instead of each bit being just 0 or 1, you can have components that represent 0 to n. But each state is still discrete, the only difference is whether representing, say, 0 to 3 requires 2 components (2 bits) or a single component (a qit?).
  - arcticbull 3y ago
    Isn't that just how MLC NAND works though? Also the article is just describing memristor memory isn't it? Intel's been shipping Optane memristor based memory (which they deny, but others are skeptical of the denial) for years, so clearly I'm missing something.
    - meling 3y ago
      Actually, Intel has discontinued Optane: https://en.wikipedia.org/wiki/3D_XPoint https://en.wikipedia.org/wiki/3D_XPoint https://www.theregister.com/2022/07/29/intel_optane_memory_dead/ https://www.theregister.com/2022/07/29/intel_optane_memory_d...
    - cubefox 3y ago
      Optane was PCM , not ReRAM (memristor).
  - noam_k 3y ago
    > ... representing, say, 0 to 3 requires 2 components (2 bits) or a single component (a qit?). I think the term you're looking for is trit (ternary/trinary digit).
    - Nursie 3y ago
      A trit is 0,1,2, no? Tri-state.
    - spiralx 3y ago
      0 to 3 would be a quat, 0 to 4 a quit, 0 to 5 a hext and so on.
- ggm 3y ago
  > Does anyone really want to process analog data? But.. you do all the time. pretty much all the time these binary states are calculated from curves, or approximations. The state isn't 1 or 0, its the set of states which make 1 or 0 depending, under error correcting codes, over-sampling, things like manchester encoding. We're fundamentally imposing binary on a non-binary state, by making deterministic judgement calls about which side of a divide things fall on, but to do that we have to look at analogue qualities. I like to imagine people who work in VLSI smile at this statement, but at some level its analogue everywhere. It's higher-states which get to act like its binary, but they map to an analogue signal substrate.
  - sydbarrett74 3y ago
    Well, analogue everywhere until quantum effects come into play.
- coldtea 3y ago
  This isn't about analog data. The states will still be quantized/digitized. Besides, currently used transistors in regular CPUs are analog too. And, does anyone want more than 0 and 1? If it's faster and has largest capacities, with equally feasible manufacturing, then yes.
- quench 3y ago
  This would still be digital (discrete) not analog. For example you can have 4 voltage levels representing 00, 01, 10, 11 This technique is used in modems. So you can store more information in each memory location resulting in higher density.
- legulere 3y ago
  There's an argument to be made, if reality isn't discrete (can be represented exactly by rational numbers) but continuous (can be represented by real numbers) you can represent programs that cannot be represented in a discrete way like a Turing machine. For instance the typical proof that the halting problem is unsolvable relies on there being just a discrete infinite amount of programs.
- freedude 3y ago
  The quality of the media is in the perception of the user. Ask any audiophile if vinyl is best or not. Analog isn't really the issue here as there would be some kind of multiplexing/demultiplexing encoding or device to interface with it. It would still be discrete and have defined levels. as the technology advances you may be able to define more levels in the same bandwidth space thus creating "more" storage by greater efficiency of the spectrum.
- kazinator 3y ago
  The encoding used on vinyl records has more information density than if zeros and ones were stored, say by two levels of the groove (or a closely related encoding like RZ or NRZ).
chimeracoder 3y ago
Binary data simply means that each datum encodes exactly one of two states. We represent them as 0s and 1s, but that's just an abstraction of convenience - physically, that maps to something like "high voltage" and "low voltage", or "dark" and "light", or "on" and "off", depending on the physical medium. We could have three states - "high voltage", "medium voltage" and "low voltage" - but the advantage of having exactly two is that it makes it harder to mistake one state for another (e.g. if voltage fluctuates within a specific range). If the measurement range for each datum is (0, 100), you can decide that anything below 50 is "low voltage" and anything below 100 is "high voltage". You can also do the same thing with (0,33), <33,66), <66,100), but that requires assuming more precise tools throughout the entire pipeline. When we talk about bits getting flipped (the entire reason that checksums exist in wire protocols), that's the reason: the medium conducting the signal is imprecise, and sometimes the reading is off. Traditionally, this represents a tradeoff between density and fidelity. If your system has high enough fidelity, you can take advantage of the additional precision and distinguish between more states, representing additional information. If your system has 8 states (0,8.3), <8.3, 16.6)... etc., you can look at this as an octal system, or you can think of it as a binary system in which a read/write error affects an entire byte, rather than a single bit. At the end of the day, this is a question of signal processing - binary representation is a convenient abstraction that allows us to understand the way that we're interpreting the signals we're reading, but it's fundamentally an arbitrary choice.
- quasse 3y ago
  This is multi-level cell flash memory, and it's been in use for years. Instead of "high" or "low" voltage they use the in-between states to encode more bits per cell. [1] I worked at an embedded startup when SD cards started switching to MLC and we saw a really notable decrease in storage reliability. We ended up sourcing special SD cards that were flashed with embedded controller firmware restoring them to single level cell functionality. Your storage space is divided by 2^n going from n levels of voltage per cell back to SLC, but we saw greatly increased data integrity. [1] https://en.wikipedia.org/wiki/Multi-level_cell https://en.wikipedia.org/wiki/Multi-level_cell
  - 3y ago
1024core 3y ago
How long have we been talking about "new" types of memory/storage? Anyone remember memristors?
- renegade-otter 3y ago
  The article literally links to another one that talks about the "emerging type of memory" - from 2012.
- dale_glass 3y ago
  The problem is that any new technology has to compete with half a century of improvement made to the current one. It's not enough to make a proof of concept of your MRAM, FRAM, or whatnot. If you want it to catch on you have to also come up with a way to quickly start making say, 16GB modules competitive with modern DDR5. That's tough.
- hanniabu 3y ago
  Has anything happened with memristors?
  - p1esk 3y ago
    With the recent successes quantizing LLMs to 4 bits, memristors can actually be useful. Basically the lower the precision the more attractive analog computation becomes.
  - mchannon 3y ago
    You can buy them at knowm.com so I'd guess.. maybe?
twic 3y ago
> Normally it’s challenging to use [hafnium oxide] for memory because it has no structure at the atomic level – its hafnium and oxygen atoms are randomly mixed together. What on earth does this mean? Looks like a typically non-random crystal to me: https://en.wikipedia.org/wiki/Hafnium(IV)_oxide https://en.wikipedia.org/wiki/Hafnium(IV)_oxide The actual article is open access, but rather dense: https://www.science.org/doi/10.1126/sciadv.adg1946 https://www.science.org/doi/10.1126/sciadv.adg1946 It seems they used amorphous hafnium oxide: > To create this amorphous nanocomposite, we added Ba to hafnium oxide during the single-step thin film deposition, and because of its simplicity of compositional control, we used pulsed laser deposition (PLD) to deposit the films. The PLD target had a Ba:Hf cation ratio of 1:2, which exceeds the solubility limit of dopants (with large atomic radii) in (crystalline) hafnium oxide, so that the formation of a second (amorphous) phase could be expected (22). Is that what the article means? Because it also then talks about adding barium on top of this, whereas you don't even get the amorphous phase without barium. Anyway, the paper also explains the deal with the Barium "bridges", which the authors call columns, fairly clearly (but it's too long to copy here). Other observations: - The authors seem to be very proud of producing this device mostly using "industry-compatible" processes, which i think means stuff you could do in a conventional chip fab, and where they don't, they have a story about how they could. This seems to contrast with other experiments in this field where the techniques would be very difficult to export from the lab. - Switching time seems to be > ~100 ns. Maybe that can be optimised a lot. And i don't know what that translates to in terms of latency at bus level. But that time is ~100x SRAM, ~30x DRAM, ~10x some of these other wacky high-speed storage technologies, but ~1000x less than flash. Could have those numbers highly mixed up, so correct me if i'm wrong. - There is some chat about "neuromorphic" functionality, where storage works like neurons. I won't go into the details but this sounds like a fad that is probably getting semiconductor researchers nice big grants, and will go absolutely nowhere.
stevefan1999 3y ago
Actually if you think in the sense of informatics...base 3 is more efficient than base 2, but natural base (base e = base 2.718...) is the most efficient, and 3 - e is closer to e than e - 2. So it is more natural for computers to go in multiple bases but unnatural to human.
- davidw 3y ago
  See: https://en.wikipedia.org/wiki/Ternary_computer https://en.wikipedia.org/wiki/Ternary_computer
  - stevefan1999 3y ago
    Yes -- The soviets did the right thing, but it was too costly.
- deleted 3y ago
  [deleted]
- aYsY4dDQ2NrcNzA 3y ago
  > 3 - e is closer to e than e - 2 I don't get it.
  - hulahoof 3y ago
    I assume they mean absolute distance abs(3-e) < abs(2-e)
    - dismalpedigree 3y ago
      I still don’t understand why base e is the most efficient ideal. There has to be some more info on this.
      - gretch 3y ago
        Yeah I don’t even understand non-whole bases. For example, how does someone express the number 5 in base e…
3cats-in-a-coat 3y ago
> As powerful as current computer technology can be, there are a few hard limits to it. Data is encoded into just two states – one or zero. And this data is stored and processed in different parts of a computer system, so it needs to be shuttled back and forth, which consumes energy and time. This kinda reads like someone who doesn't know how storage works. Storage is rarely encoded as "one and zero" in the form you see it, say, in a hex editor. In fact that's what's it decoded to. Storage schemes vary, to limit run length, add error correction, and in some cases includes analog states like the ones described in the article, to make more efficient use of charge and material state. This is trivial, because actually the smallest unit you work with in a computer is not a bit, but a byte. And often in fact a word, which is typically 16 bytes (64 bits), or 64 bytes (512 bits) the size of a cache line, or even 4096 bytes, the size of a typical memory page or disk sector. So we have plenty of leeway to comfortably encode these larger units however we wish, and the rest can stay as-is.
kens 3y ago
The Intel 8087 math coprocessor (1980) had too much microcode to fit into a standard ROM. It was implemented with a special ROM that stored two bits per transistor by using four different transistor sizes, almost doubling the density. Each value read from the ROM was thresholded to generate two output bits. This technique was also used in the iAPX 432's I/O processor chip.
- yonatan8070 3y ago
  Isn't this exactly what modern TLC/QLC do?
  - colejohnson66 3y ago
    Same concept for reading - threshold regions indicate multiple bit values, but different for storage. The 8087 used physically different transistor sizes[0] to read a voltage level that "stored" two bits inside four threshold regions. MLC/TLC/QLC/PLC use a single transistor but charge the gate to a different "analog" voltage that is between 2^n threshold regions - MLC has four (two bits), TLC has eight (three bits), QLC has 16 (four bits), and PLC has 32 (five bits). [0]: https://www.righto.com/2018/09/two-bits-per-transistor-high-density.html https://www.righto.com/2018/09/two-bits-per-transistor-high-...
- formerly_proven 3y ago
  NB-ROM-shenanigans are very traditional. The Tektronix 7000 readout (1969) managed to encode the stroke of an entire letter with just a few transistors. Here's the letter 3, encoded in 24 transistors: https://w140.com/tekwiki/images/e/e0/7000-readout-chargen-circuit-1.jpg https://w140.com/tekwiki/images/e/e0/7000-readout-chargen-ci... This uses multi-emitter transistors to encode discrete current ratios (which are then interpolated across groups by smoothly ramping base currents between points, turning them into strokes), all connections done in the metal layer, so the silicon was the same for all ROM chips. https://w140.com/tekwiki/wiki/7000_series_readout_system https://w140.com/tekwiki/wiki/7000_series_readout_system
  - jojobas 3y ago
dcow 3y ago
Ah, memristors. I didn't know term had fallen out of popularity. These things were all the rage up till about 10 years ago. “The 4th fundamental missing electrical circuit component”: resistor, inductor, capacitor, and.. memristor. We were supposed to build hardware neural nets the size of human brains with these things, and get petabytes of storage in unheard-of densities. What happened?
- _visgean 3y ago
  but this is not memristor afaik, the breakthrough here is that it has multiple states, not that it can remember the states without power. (At least the article does not mention if the state is preserved so I would assume not.)
  - dcow 3y ago
    No it has a continuum of states. It’s the same thing.
nomercy400 3y ago
0 and 1 are human definitions like time and date. You can define any state to be a 0 and any other to be 1, whether it is flowing electrons, flowing water, on-off light, etc. By doing this you gain a simpler model on which you can reason. You can even expand it to whether a 'neuron' is firing (1) or not (0). What causes the firing does not matter, may it be digital or analog, discrete or continuous. You want to represent more than two states? Use multiple 1s or 0s, and you can still use the model. Use something else? Now you have to create your own model where you can map your states and state changes, along with all mathematics associated. You can no longer compare a USB 'byte' with this new model 'byte'. What even is a 'byte' in your new model? It's like comparing apples and oranges. It usually all depends on how accurately you can measure your states, often limited by the signal-to-noise ratio. When is a 0 really a 0 (signal) and not a 1 (noise). A good example is telecommunications, where the data is sent over an electromagnetic wave (analog, continuous). In the past you could only identify between 32 states, but now due to better sensors you can identify 2048 states, increasing your data density by 64x. Same goes for HDD and SD card storage.
p0w3n3d 3y ago
Bender: "What an awful dream! 1s and 0s everywhere! (gasp) I thought I saw 2..."
aaron695 3y ago
[dead]
nycdotnet 3y ago
Who could have predicted that the key element to break the hegemony of 0 and 1 is ha(l)fnium.

Vinyl records also don’t have 0’s and zeroes, it does not make them better than Blu-ray. MLC NAND have more than just 1’s and 0’s, they have 0..7 or even 0..15. Does anyone really want to process analog data?

vlovich123 3y ago
> Does anyone really want to process analog data? I imagine at some point in the abstractions it'll get a compat layer. Also, I could be wrong about MLC NAND but I thought it still stored 0/1 and just had more bits stored per cell through stacking.
- whatevaa 3y ago
  No, multi-level NAND has extra voltage levels. Instead of a just regular off (0) and on (1) they have intermediate states too. For QLC, that's 16 levels, it's quite granular, and the margins of error are very small.
  - Retr0id 3y ago
    The margins for error are wider than one might immediately assume, since they combine it with error correction codes.
- sp332 3y ago
  TLC NAND stores one of eight values in the cell. There aren't three separate values like 1, 0, 1, there is only one value, like "5".
  - AussieWog93 3y ago
    I mean, 101 is just 5 in binary. So each cell stores either three pieces of binary or one piece of octal. It's the same thing.
    - failuser 3y ago
      5 is not stored as 101, it’s stored as charge level number 5 out of 8 other valid charge levels.
    - koromak 3y ago
      Its absolutely not the same thing from a hardware perspective. Normal components always operate on bits. We see "5", but the computer sees (at least) 3 bits. It then sends its output as bits to the next one. A 5 must require 3 wires, or 3 cycles of the clock over a single wire. Building a computer where each wire can encode multiple bits of data at the exact same time fundamentally changes the way operations are done. The entire thing would need to be built in a different way, with different consequences. It might be a bad idea, or impossible, but it would be different.
mcdonje 3y ago
1. Vinyl ridges are much larger than DVD lines. Kind of a silly comparison considering the article claims the density of this scheme would exceed what we currently have. 2. It appears to not be analog. >An energy barrier is created at the points where the bridges meet the device contacts, and the height of this barrier can be controlled which changes the electrical resistance of the overall material. That in turn is what encodes the data. That seems to indicate they can have a base-n encoding where n is what they can practically achieve with the barrier/bridge layering scheme.
- failuser 3y ago
  They claim it’s continuous. If continuous values itself were that great then the vinyl records were great for data storage. The problem with continuous values is noise, durability and precision. You can made vinyl disks with incredibly high precision, but preventing damage and reading back those values is not practical. That’s why even the first analog laser medium had discrete pits and lands instead of a continuous signal.
  - AnotherGoodName 3y ago
    It can be continuous but still have a limit to data due to noise. In vinyls case the more data you store the higher the error rate (at some point the differences in states is hidden in noise). The data rate from a noisy channel or storage can be derived from Shannon-hartley theorem. This is why the old analogue phones could only do 56k for example.
    - failuser 3y ago
      Yes, that’s why I hate their headline.
    - coldtea 3y ago
      >This is why the old analogue phones could only do 56k for example. And somehow, on analogue circuit switch phone call mode, still sounded better with less artifacts than most mobile phone calls!
PhasmaFelis 3y ago
> Does anyone really want to process analog data? If they can get it to encode binary data more densely than existing systems, with acceptable speed/reliability/cost? Yes, of course they would.
kstrauser 3y ago
My understanding is that modern hard drives all use probability and error detection/correction to convert a gooey analog signal into chunky discrete goodness.
dragontamer 3y ago
> Does anyone really want to process analog data? I mean... maybe we should? All radio waves, such as 802.11, are fundamentally analog data. And then they are converted back into digital with incredible QAM-encoding levels. If you think QLC 4-bits per Flash-cell is big, wait till you see 256-QAM (aka: 8-bits per 802.11 timeslice), or more. I believe I saw some spec for some fiber-optic cable that is 32768-QAM (aka: 15-bits per timeslice). -------- Then again, Flash (and other memory) are composed of individual cells. True-analog signals like radio-waves are truly analog, so maybe its not quite the same. But maybe we should think of things like tape or hard-drives as an analog signal. Or maybe not, who knows? I'm not actually a storage specialist so I don't really know. (Nor am I really in communications, lol).
- foobiekr 3y ago
  256+ QAM is there but you will almost never see it in practice unless you’re next to the AP. Wifi7 defines up to 4096 so maybe the economics will drive improvements that didn’t land in 6E APs.
  - tguvot 3y ago
    docsis. up to 16384
  - Taniwha 3y ago
    You'll see it on almost every cableTV channel these days
    - foobiekr 3y ago
      I was referring to Wifi but yes
- enkid 3y ago
  I'm pretty sure that Einstein proved that light (including radio) is not, in-fact, analog, but rather has discrete levels of energy in quantum steps.
hn_throwaway_99 3y ago
I think this is a faulty analogy. The data is still encoded (or, at least, interpreted) as digital data, except that instead of each bit being just 0 or 1, you can have components that represent 0 to n. But each state is still discrete, the only difference is whether representing, say, 0 to 3 requires 2 components (2 bits) or a single component (a qit?).
- arcticbull 3y ago
  Isn't that just how MLC NAND works though? Also the article is just describing memristor memory isn't it? Intel's been shipping Optane memristor based memory (which they deny, but others are skeptical of the denial) for years, so clearly I'm missing something.
  - meling 3y ago
    Actually, Intel has discontinued Optane: https://en.wikipedia.org/wiki/3D_XPoint https://en.wikipedia.org/wiki/3D_XPoint https://www.theregister.com/2022/07/29/intel_optane_memory_dead/ https://www.theregister.com/2022/07/29/intel_optane_memory_d...
  - cubefox 3y ago
    Optane was PCM , not ReRAM (memristor).
- noam_k 3y ago
  > ... representing, say, 0 to 3 requires 2 components (2 bits) or a single component (a qit?). I think the term you're looking for is trit (ternary/trinary digit).
  - Nursie 3y ago
    A trit is 0,1,2, no? Tri-state.
  - spiralx 3y ago
    0 to 3 would be a quat, 0 to 4 a quit, 0 to 5 a hext and so on.
ggm 3y ago
> Does anyone really want to process analog data? But.. you do all the time. pretty much all the time these binary states are calculated from curves, or approximations. The state isn't 1 or 0, its the set of states which make 1 or 0 depending, under error correcting codes, over-sampling, things like manchester encoding. We're fundamentally imposing binary on a non-binary state, by making deterministic judgement calls about which side of a divide things fall on, but to do that we have to look at analogue qualities. I like to imagine people who work in VLSI smile at this statement, but at some level its analogue everywhere. It's higher-states which get to act like its binary, but they map to an analogue signal substrate.
- sydbarrett74 3y ago
  Well, analogue everywhere until quantum effects come into play.
coldtea 3y ago
This isn't about analog data. The states will still be quantized/digitized. Besides, currently used transistors in regular CPUs are analog too. And, does anyone want more than 0 and 1? If it's faster and has largest capacities, with equally feasible manufacturing, then yes.
quench 3y ago
This would still be digital (discrete) not analog. For example you can have 4 voltage levels representing 00, 01, 10, 11 This technique is used in modems. So you can store more information in each memory location resulting in higher density.
legulere 3y ago
There's an argument to be made, if reality isn't discrete (can be represented exactly by rational numbers) but continuous (can be represented by real numbers) you can represent programs that cannot be represented in a discrete way like a Turing machine. For instance the typical proof that the halting problem is unsolvable relies on there being just a discrete infinite amount of programs.
freedude 3y ago
The quality of the media is in the perception of the user. Ask any audiophile if vinyl is best or not. Analog isn't really the issue here as there would be some kind of multiplexing/demultiplexing encoding or device to interface with it. It would still be discrete and have defined levels. as the technology advances you may be able to define more levels in the same bandwidth space thus creating "more" storage by greater efficiency of the spectrum.
kazinator 3y ago
The encoding used on vinyl records has more information density than if zeros and ones were stored, say by two levels of the groove (or a closely related encoding like RZ or NRZ).

chimeracoder 3y ago

Binary data simply means that each datum encodes exactly one of two states. We represent them as 0s and 1s, but that's just an abstraction of convenience - physically, that maps to something like "high voltage" and "low voltage", or "dark" and "light", or "on" and "off", depending on the physical medium. We could have three states - "high voltage", "medium voltage" and "low voltage" - but the advantage of having exactly two is that it makes it harder to mistake one state for another (e.g. if voltage fluctuates within a specific range). If the measurement range for each datum is (0, 100), you can decide that anything below 50 is "low voltage" and anything below 100 is "high voltage". You can also do the same thing with (0,33), <33,66), <66,100), but that requires assuming more precise tools throughout the entire pipeline. When we talk about bits getting flipped (the entire reason that checksums exist in wire protocols), that's the reason: the medium conducting the signal is imprecise, and sometimes the reading is off. Traditionally, this represents a tradeoff between density and fidelity. If your system has high enough fidelity, you can take advantage of the additional precision and distinguish between more states, representing additional information. If your system has 8 states (0,8.3), <8.3, 16.6)... etc., you can look at this as an octal system, or you can think of it as a binary system in which a read/write error affects an entire byte, rather than a single bit. At the end of the day, this is a question of signal processing - binary representation is a convenient abstraction that allows us to understand the way that we're interpreting the signals we're reading, but it's fundamentally an arbitrary choice.

quasse 3y ago
This is multi-level cell flash memory, and it's been in use for years. Instead of "high" or "low" voltage they use the in-between states to encode more bits per cell. [1] I worked at an embedded startup when SD cards started switching to MLC and we saw a really notable decrease in storage reliability. We ended up sourcing special SD cards that were flashed with embedded controller firmware restoring them to single level cell functionality. Your storage space is divided by 2^n going from n levels of voltage per cell back to SLC, but we saw greatly increased data integrity. [1] https://en.wikipedia.org/wiki/Multi-level_cell https://en.wikipedia.org/wiki/Multi-level_cell
- 3y ago

1024core 3y ago

How long have we been talking about "new" types of memory/storage? Anyone remember memristors?

renegade-otter 3y ago
The article literally links to another one that talks about the "emerging type of memory" - from 2012.
dale_glass 3y ago
The problem is that any new technology has to compete with half a century of improvement made to the current one. It's not enough to make a proof of concept of your MRAM, FRAM, or whatnot. If you want it to catch on you have to also come up with a way to quickly start making say, 16GB modules competitive with modern DDR5. That's tough.
hanniabu 3y ago
Has anything happened with memristors?
- p1esk 3y ago
  With the recent successes quantizing LLMs to 4 bits, memristors can actually be useful. Basically the lower the precision the more attractive analog computation becomes.
- mchannon 3y ago
  You can buy them at knowm.com so I'd guess.. maybe?

twic 3y ago

> Normally it’s challenging to use [hafnium oxide] for memory because it has no structure at the atomic level – its hafnium and oxygen atoms are randomly mixed together. What on earth does this mean? Looks like a typically non-random crystal to me: https://en.wikipedia.org/wiki/Hafnium(IV)_oxide https://en.wikipedia.org/wiki/Hafnium(IV)_oxide The actual article is open access, but rather dense: https://www.science.org/doi/10.1126/sciadv.adg1946 https://www.science.org/doi/10.1126/sciadv.adg1946 It seems they used amorphous hafnium oxide: > To create this amorphous nanocomposite, we added Ba to hafnium oxide during the single-step thin film deposition, and because of its simplicity of compositional control, we used pulsed laser deposition (PLD) to deposit the films. The PLD target had a Ba:Hf cation ratio of 1:2, which exceeds the solubility limit of dopants (with large atomic radii) in (crystalline) hafnium oxide, so that the formation of a second (amorphous) phase could be expected (22). Is that what the article means? Because it also then talks about adding barium on top of this, whereas you don't even get the amorphous phase without barium. Anyway, the paper also explains the deal with the Barium "bridges", which the authors call columns, fairly clearly (but it's too long to copy here). Other observations: - The authors seem to be very proud of producing this device mostly using "industry-compatible" processes, which i think means stuff you could do in a conventional chip fab, and where they don't, they have a story about how they could. This seems to contrast with other experiments in this field where the techniques would be very difficult to export from the lab. - Switching time seems to be > ~100 ns. Maybe that can be optimised a lot. And i don't know what that translates to in terms of latency at bus level. But that time is ~100x SRAM, ~30x DRAM, ~10x some of these other wacky high-speed storage technologies, but ~1000x less than flash. Could have those numbers highly mixed up, so correct me if i'm wrong. - There is some chat about "neuromorphic" functionality, where storage works like neurons. I won't go into the details but this sounds like a fad that is probably getting semiconductor researchers nice big grants, and will go absolutely nowhere.

stevefan1999 3y ago

Actually if you think in the sense of informatics...base 3 is more efficient than base 2, but natural base (base e = base 2.718...) is the most efficient, and 3 - e is closer to e than e - 2. So it is more natural for computers to go in multiple bases but unnatural to human.

davidw 3y ago
See: https://en.wikipedia.org/wiki/Ternary_computer https://en.wikipedia.org/wiki/Ternary_computer
- stevefan1999 3y ago
  Yes -- The soviets did the right thing, but it was too costly.
deleted 3y ago
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aYsY4dDQ2NrcNzA 3y ago
> 3 - e is closer to e than e - 2 I don't get it.
- hulahoof 3y ago
  I assume they mean absolute distance abs(3-e) < abs(2-e)
  - dismalpedigree 3y ago
    I still don’t understand why base e is the most efficient ideal. There has to be some more info on this.
    - gretch 3y ago
      Yeah I don’t even understand non-whole bases. For example, how does someone express the number 5 in base e…

3cats-in-a-coat 3y ago

> As powerful as current computer technology can be, there are a few hard limits to it. Data is encoded into just two states – one or zero. And this data is stored and processed in different parts of a computer system, so it needs to be shuttled back and forth, which consumes energy and time. This kinda reads like someone who doesn't know how storage works. Storage is rarely encoded as "one and zero" in the form you see it, say, in a hex editor. In fact that's what's it decoded to. Storage schemes vary, to limit run length, add error correction, and in some cases includes analog states like the ones described in the article, to make more efficient use of charge and material state. This is trivial, because actually the smallest unit you work with in a computer is not a bit, but a byte. And often in fact a word, which is typically 16 bytes (64 bits), or 64 bytes (512 bits) the size of a cache line, or even 4096 bytes, the size of a typical memory page or disk sector. So we have plenty of leeway to comfortably encode these larger units however we wish, and the rest can stay as-is.

kens 3y ago

The Intel 8087 math coprocessor (1980) had too much microcode to fit into a standard ROM. It was implemented with a special ROM that stored two bits per transistor by using four different transistor sizes, almost doubling the density. Each value read from the ROM was thresholded to generate two output bits. This technique was also used in the iAPX 432's I/O processor chip.

yonatan8070 3y ago
Isn't this exactly what modern TLC/QLC do?
- colejohnson66 3y ago
  Same concept for reading - threshold regions indicate multiple bit values, but different for storage. The 8087 used physically different transistor sizes[0] to read a voltage level that "stored" two bits inside four threshold regions. MLC/TLC/QLC/PLC use a single transistor but charge the gate to a different "analog" voltage that is between 2^n threshold regions - MLC has four (two bits), TLC has eight (three bits), QLC has 16 (four bits), and PLC has 32 (five bits). [0]: https://www.righto.com/2018/09/two-bits-per-transistor-high-density.html https://www.righto.com/2018/09/two-bits-per-transistor-high-...
formerly_proven 3y ago
NB-ROM-shenanigans are very traditional. The Tektronix 7000 readout (1969) managed to encode the stroke of an entire letter with just a few transistors. Here's the letter 3, encoded in 24 transistors: https://w140.com/tekwiki/images/e/e0/7000-readout-chargen-circuit-1.jpg https://w140.com/tekwiki/images/e/e0/7000-readout-chargen-ci... This uses multi-emitter transistors to encode discrete current ratios (which are then interpolated across groups by smoothly ramping base currents between points, turning them into strokes), all connections done in the metal layer, so the silicon was the same for all ROM chips. https://w140.com/tekwiki/wiki/7000_series_readout_system https://w140.com/tekwiki/wiki/7000_series_readout_system
- jojobas 3y ago

dcow 3y ago

Ah, memristors. I didn't know term had fallen out of popularity. These things were all the rage up till about 10 years ago. “The 4th fundamental missing electrical circuit component”: resistor, inductor, capacitor, and.. memristor. We were supposed to build hardware neural nets the size of human brains with these things, and get petabytes of storage in unheard-of densities. What happened?

_visgean 3y ago
but this is not memristor afaik, the breakthrough here is that it has multiple states, not that it can remember the states without power. (At least the article does not mention if the state is preserved so I would assume not.)
- dcow 3y ago
  No it has a continuum of states. It’s the same thing.

nomercy400 3y ago

0 and 1 are human definitions like time and date. You can define any state to be a 0 and any other to be 1, whether it is flowing electrons, flowing water, on-off light, etc. By doing this you gain a simpler model on which you can reason. You can even expand it to whether a 'neuron' is firing (1) or not (0). What causes the firing does not matter, may it be digital or analog, discrete or continuous. You want to represent more than two states? Use multiple 1s or 0s, and you can still use the model. Use something else? Now you have to create your own model where you can map your states and state changes, along with all mathematics associated. You can no longer compare a USB 'byte' with this new model 'byte'. What even is a 'byte' in your new model? It's like comparing apples and oranges. It usually all depends on how accurately you can measure your states, often limited by the signal-to-noise ratio. When is a 0 really a 0 (signal) and not a 1 (noise). A good example is telecommunications, where the data is sent over an electromagnetic wave (analog, continuous). In the past you could only identify between 32 states, but now due to better sensors you can identify 2048 states, increasing your data density by 64x. Same goes for HDD and SD card storage.

p0w3n3d 3y ago

Bender: "What an awful dream! 1s and 0s everywhere! (gasp) I thought I saw 2..."

aaron695 3y ago

[dead]

nycdotnet 3y ago

Who could have predicted that the key element to break the hegemony of 0 and 1 is ha(l)fnium.