4 ms·
Limits of computation
- prashnts 9y ago> A cold degenerate star could conceivably be used as a giant data storage device, [...] purposes. The cited link to this [1] is a fascinating email thread/essay which they call "Femtotech". I don't even understand most of the stuff, but it seems we're far, far from it. [1] https://web.archive.org/web/20041025030505/http://www.cs.usu.edu/~degaris/essays/femtotech.html https://web.archive.org/web/20041025030505/http://www.cs.usu...
- projectorlochsa 9y agoI remember watching some documentaries and reading some articles, and remember Hugo de Garis talking about femtotech. I thought he was a pioneer in the technology, seems he's just a bystander. He and Ben Goertzel seem to look as pioneers but never produce something tangible.
- deleted 9y ago[deleted]
- pmontra 9y agoAbout reversible computation and the Landauer limit: http://www.nipslab.org/node/5097 http://www.nipslab.org/node/5097
- agumonkey 9y agoI found out about Landauer principe last week and am fascinated about the mental shift.
- vilhelm_s 9y agoI think the key paragraph from the paper is this: > We would like to emphasize that the output of our logic gate can be fed directly as an input to a second similar device (as shown below) and so on, to perform all the calculations desired. Clearly, our logic gate is a combinational device; thus, the removal of the inputs makes the system revert to the initial state S0 (regardless of whether the final state was S0 or S1). If we want to remember the final state, we need to couple this device to a sequential device where a Landauer reset might be required and a minimum dissipation of kBT log 2 is needed. In other words, the Laundauer limit applies to state machines, but not to combinatorial logic blocks. I guess in retrospect this should not be surprising: the derivation of the principle considers a computing machine which is in thermal equilibrium with its environment, so implicitly we model the machine as having some memory unit which has time to thermalize after each operation.
- smeyer 9y ago>Landauer's principle defines a lower theoretical limit for energy consumption: kT ln 2 joules consumed per irreversible state change, where k is the Boltzmann constant and T is the operating temperature of the computer.[2] Reversible computing is not subject to this lower bound. T cannot, even in theory, be made lower than 3 kelvins, the approximate temperature of the cosmic microwave background radiation, without spending more energy on cooling than is saved in computation. Well, technically, you just have to wait as the universe continues expanding if you want the temperature of the CMB to drop.
- jakeogh 9y agoBlack Hole Farming: https://www.youtube.com/watch?v=Qam5BkXIEhQ https://www.youtube.com/watch?v=Qam5BkXIEhQ
- rorosaurus 9y agoFascinating analysis! The section "Building devices that approach physical limits" reminds me of a rock Kurzweil mentions in "The Singularity is Near" [1]. Although it's not as a cosmic scale, it's a similarly interesting exploration of the limits of computation! >“How Smart Is a Rock? To appreciate the feasibility of computing with no energy and no heat, consider the computation that takes place in an ordinary rock. Although it may appear that nothing much is going on inside a rock, the approximately 1025 (ten trillion trillion) atoms in a kilogram of matter are actually extremely active. Despite the apparent solidity of the object, the atoms are all in motion, sharing electrons back and forth, changing particle spins, and generating rapidly moving electromagnetic fields. All of this activity represents computation, even if not very meaningfully organized. We’ve already shown that atoms can store information at a density of greater than one bit per atom, such as in computing systems built from nuclear magnetic-resonance devices. University of Oklahoma researchers stored 1,024 bits in the magnetic interactions of the protons of a single molecule containing nineteen hydrogen atoms.51 Thus, the state of the rock at any one moment represents at least 1027 bits of memory.” [1] https://www.goodreads.com/quotes/1284270-how-smart-is-a-rock-to-appreciate-the-feasibility-of https://www.goodreads.com/quotes/1284270-how-smart-is-a-rock...
- philipkglass 9y agoThat seems to deeply confuse "a thing that can be modeled by computer" with "a computer." It's trying to commensurate the incommensurable. How much more efficient is a rock than your phone when it comes to computing the sum of two 64 bit integers? How much more efficient when it comes to storing and retrieving a megabit of JPEG data? I'd accept answers in the form of joules per operation. The very questions are faulty. A factory-stock rock is not a programable logic or memory device. If it computes at all, it only computes being-a-rock. This rock analogy is the sort of sloppy thinking that takes someone from "it takes at least 38 petaflops to simulate a conscious human brain" (currently unknown, but not obviously incorrect) to "the human brain is a petaflop-class supercomputer" (WTF, no, category error).
- the8472 9y agoIt's basically saying that the universe is a special kind of computer that computes state transitions according to the rules of physics, presumably at the planck scale, which is far beyond what our simplified models of said rules can manage. This view of he universe seems to be fairly important and taken seriously since black hole thermodynamics, gravity as entropic force and similar concepts are derived from it.
- kangxitenant 9y agohttps://arxiv.org/abs/1301.4504 https://arxiv.org/abs/1301.4504 quantum computation complexity prevents hawking radiation from being decoded before the black hole evaporates!
- dorgo 9y agoThere are noncomputable functions. For example: https://en.wikipedia.org/wiki/Busy_beaver https://en.wikipedia.org/wiki/Busy_beaver
- csl 9y agoIf anyone is interested, I attempted to plot the first few Busy Beaver iterations: https://github.com/cslarsen/busy-beaver https://github.com/cslarsen/busy-beaver
- dorgo 9y ago>Busy beaver machines are, by definition, non-halting machines Is this an error? I thought busy beaver must halt?
- csl 9y agoThanks, I see now that the whole section is very clumsily written. IIRC, a Busy Beaver may halt or not. You only care about those that _do_ halt, though, and the _champion_ is the one with the most number of 1s on the tape. The whole problem is determining if a given machine will halt or not. Which we know, by the halting theorem, is possible to prove on a case-by-case basis, but not in the general case (i.e., you cannot create an algorithm to determine it). So you resort to heuristics. E.g., if a machine never has a "go to the halt state" in its transition table, you know that it can't ever halt. But as the machines grow, you'll have to cover an infinite amount of such heuristics, hence the incomputable nature of them.