In contrast, when arguing that the timeline 'splits' due to measurements, the resulting universes do not interact at all and remain completely unaware of each other—they can never even know if the others exist.
If quantum computers truly 'forked' the world, they would be equivalent to non-deterministic Turing machines (capable of solving NP-complete problems in polynomial time), but quantum computing experts agree that they can still be modeled as deterministic Turing machines.
It's better to think of quantum computers as a type of analog computer, capable of solving certain problems that fit their model well, but not generally more powerful. It’s like an Intel CPU having SIMD or AVX instructions that allow it to perform certain operations faster, but these don't fundamentally change its capabilities. The no-free-lunch theorem applies.
A typical quantum algorithm like Shor's works by sending every possible input through a gate, and so you get every possible output out in a superposition. If you were to just measure that, you'd get a random result - so instead, you need to somehow interfere the output to get the actual result. You do this by taking advantage of the fact that the superposition is a periodic function and the amplitude repeats. This is literally the core assumption of the algorithm.(a common way of doing this using the QFT).
Every quantum algorithm requires some kind of structure in the output like this. Deustch's algo, dumb ones like Simon's algo, etc. NP-Complete problems have no structure to them, so even if you build a gate that creates the superposition you want, it's not possible to destructively interfere it to get an answer (I don't know how to prove that there's no structure to NP-Complete outputs - it just feels trivial, since they're only solvable in exponential time, so there must be an exponential amount of "structure" there).
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[1] The only way to communicate with the other universe would be to try to use quantum mechanics with something like an entangled pair. But no information can be communicated through an entangled pair if all you just have 1 of the 2 particles! Measurement collapses a state nonlocally, and if you could somehow measure one particle and change the probability distribution of the other, you'd be communicating faster than light. The measurement genuinely changes the state and the amplitudes, but not in a way that the other person can detect. It's really interesting and leads to stuff like teleportation.
i.e. BPP is contained in BQP but the converse is thought to be false.
Why is there a speedup in quantum, though? Why can't you just brute force classically? The answer is that whether quantum or classical, you can always build a hard-coded circuit that essentially swaps the time and space complexity - just make it so that for every operation you were doing in time, instead, every operation happens at its own place in space.
Quantum is special because it also takes the "log" of the space complexity b/c n qubits represent 2^n bits. So quantum lets you swap space with time and then take the log of time, lol. Superposition, interference, etc aren't really even needed in the explanation.
Okay team, we’ve effectively entangled the success of our endeavor with the quantum dead man’s switch by all swearing to comply with the protocol. It’s time to start letting the universe tell us what works. QUESTION 1 for the Profit Manifold: promote yours truly to director or stay the course? Click bang hiss: 01.
Note to self: cut “universe B” (or just B? It’d hurt less) into my thigh with a razor blade 6 months before demonstrating The Device, as a plot device to be exploited for purposes TBD.
what difference is there to any 50/50 choice mechanism you chose, other than being horrendously expensive to implement?
Statement: I will make a minimal BTC transfer today, May 11. 2024, as near noon UTC+2 if the outcome is 1/true.
Result: [ { "0": 0.5140027374683648, "1": 0.4859972625316353 } ] So now we have(will) also split the bitcoin hash forever in more than 2 different branches (I guess way more than 2).
@andrewp123 Check my blog post. https://initsix.dev/in-finite-central-curve/ ,feedback appreciated :D
Morty: Aw, geez, Rick, I don't think we ought to--
Rick: Nothing ventured, nothing gained, Morty. Let's goooooooooooo!
[He clicks the mouse, and Solitaire comes up]
Rick: Haha, nice. [He starts playing.]
[We are shown a 16x16 grid of the same moment happening in 256 alternate universes.]
Ricks: Nothing ventured, nothing gained, Morty. Let's goooooooooooo!
[They click the mouse.]
Rick 183: Oh, sh--
Rick 39: Jesus Chr--
Rick 201: NO! We've gone too fa--
[237 of the alternate universes disappear in a white-hot light, their squares replaced by static.]
This is not true. It only provably does not exist in local hidden variables.
In actual practice there might as well be hidden local variables here. You wouldn't be able to tell the difference, even though you could in theory.
My preferred interpretation:
There is a density function across all possible realities (Hilbert space).
Schrodinger's cat has equal density of being alive and dead.
The person who opens the box can be happy or sad.
The density of cat being alive is entangled with the observer being happy. And the opposite for the death.
The original cat distribution did not "collapse" or "resolve" per se. The cat is still equal parts alive and dead. But it did become non-uniformly entangled with the distributions of rest of the universe.
Perhaps this is the many worlds interpretation.
probably a sign there is no real discussion here :/
It's theoretically useful, but quantum random number generation has been around for a long time. All you need is a way to detect nuclear decay.
Don’t worry though, even the professional researchers I’ve worked with think it’s a waste of time. The field is screwed.
Here’s a quick explanation from me- The state |x> means you have some qubits that represent the number x. Say you want to represent the number 13, that just means you have |1,0,1,1>, it just means you have 4 qubits in this configuration (quits can be 0 or 1). It’s also written |13>. If you want the state “13 AND 14 AND 15” in superposition where qubits are both 0 and 1, that’s represented by |1,0,1,1> + |1,1,0,0> + |1,1,0,1>. It’s in that superposition and can interact with itself until you choose to measure it. When you do go to measure it, you might measure any of the values (you dont get to choose which). Maybe you measure 15, that means the state is now |1,1,0,1>, you just deleted all the terms that aren’t 15.
This is a full pic of Shor’s algorithm https://images.app.goo.gl/ZE5rDxHScm4LUqms6
If you look at the pic, main idea is the first layer of H’s creates the state sum_x=0…2^n-1 |x, 0>, then gate U turns that state into sum_x |x, f(x)>, then the measurements measure which f(x) you have, deleting all the terms that don’t have that f(x) in them, so for example if you measure that f(x) is 13, the state is now |0, 13> + |15, 13> + |30, 13> + |45, 13> + … This is the periodic state. Now that we have it we can just apply a gate that takes the QFT (finds the frequency, which here turns the state into roughly |15, 13>), and then measures it, giving the answer period=15.
[1] https://www.pbs.org/show/pbs-space-time/
[2] https://www.youtube.com/channel/UC7_gcs09iThXybpVgjHZ_7g
How about this, they can't do the thing NP stands for. They can't run a generic polynomial-time algorithm in a nondeterministically-branching way, and then pick the winner.
I do not see the distinction between what you have written in the second paragraph and the claim that P≠NP.
Because it is deterministic, that means it's using a different algorithm than the straightforward nondeterministic solution.
So any Turing machine can solve the problem, but it will have to use the former algorithm. It can't use the latter algorithm.
A lot of people think of quantum computers as basically a nondeterministic Turing machine, and the wording in the parent post is an attempt to correct that misconception.
I still don’t understand what is meant by ’straightforward nondeterministic solution’. If you mean that QTMs or quantum circuits aren’t formally the same as NTMs, I agree. But nor are two-tape (D)TMs and single-tape DTMs, and yet we have linear speedup. Presumably there is a stronger claim here, but I do not understand what that is.
I'm not sure how much better I can phrase it than "run a generic polynomial-time algorithm in a nondeterministically-branching way, and then pick the winner."
Let me lay out a scenario:
You have a problem with 2^n potential answers.
You have a fast polynomial algorithm you can apply to each potential answer, and you want to find the right answer.
A lot of people think that if you use a quantum computer, it will check every answer in parallel and quickly find the answer at polynomial speed.
But it doesn't work like that. To some extent it does check every answer in parallel, but you need to do a lot more to make the correct answer show up out of the noise. Let's say you need to use Grover's algorithm. That cuts 2^n down to 2^(n/2), which is a remarkable speedup but it's still exponential.
It's possible that P=NP and there's a totally different solution for both classical and quantum computers that isn't exponential. But that's a side issue. The important thing here is that it's not being quantum that lets you escape exponential purgatory.
> If you mean that QTMs or quantum circuits aren’t formally the same as NTMs, I agree.
Kind of?
But it's not really about formal equivalences. It's that tons of people think a QTM is literally a NTM.