Wish I saw visualizations like these before I had to tackle the subjects. Very well done, thank you!
@jaccordero41888 сағат бұрын
Every time they say 'balls' and 'pair of balls' makes me go 💀
@wtf_gws012 күн бұрын
Found helpful 🩶
@detirei2 күн бұрын
Bogosort made this video too long...
@LittleFrog43 күн бұрын
There should be a paradox named after x lol
@philandthai3 күн бұрын
Here is the thing. I’m 73 and I grew up reading books (you sprogs can Google what they are) so I would really, really some text references.
@ascuolaconale55375 күн бұрын
What's the music playing?
@amirhosein5677y5 күн бұрын
it was awesome thanks a lot
@mrxzero50995 күн бұрын
Please upload the next videos in the series, that is the best explanation i have ever seen, very fun and includes all possible information , thank you for your hard work <3333333
@notfoundle5 күн бұрын
Look what I have found! This is the most terrific video that explains quantum entanglement I've ever seen!
@khvediri5 күн бұрын
Where does the sin^2(θ/2) and cos^2(θ/2) come from?
@randyzeitman13546 күн бұрын
Is a tensor a local taxonomy of an invarient instance?
@mikajacquel6 күн бұрын
Les parties au piano sont très sympas, qqn aurait une partition ? 😊
@ThorOdinsson297 күн бұрын
Incredible. I finally understand how tides work.
@HarKSeven8 күн бұрын
This explanation is amazing!
@XNorYT10 күн бұрын
A machine cannot solve the halting problem for another machine that contains it.
@chaumas8 күн бұрын
How do you define “contains it”?
@XNorYT8 күн бұрын
@chaumas Contains a machine that has an effect on any of its logic.
@chaumas8 күн бұрын
@ _Any_ of its logic? That would be all machines. Which is false, of course, because a machine can solve the halting problem for certain subsets of machines.
@XNorYT8 күн бұрын
@chaumas Okay, that was a bit too loosely worded. My point is that no machine can solve the halting problem with a definite and correct answer if it is requested to solve the problem with a compound machine that uses it in any of its processing.
@chaumas7 күн бұрын
@@XNorYT The problem is that unless you can define what you mean by that precisely, there’s not a lot we can say about it. However, just going by an intuitive interpretation, say we build a machine like X, but we replace N with a machine that doesn’t negate, and instead just does whatever H says. Now _any_ implementation of H that gives either answer will always correctly answer the halting problem for this case, even though this machine contains a full on copy of H. So your statement, just going by an intuitive interpretation doesn’t seem to be true.
@fasty9310 күн бұрын
In simple terms. Remember the phrase "This phrase is false". In order for H to be correct, you would have to get rid of N. H's job is to say what X will say, but in order for H to say the right thing, he needs to say the wrong thing at the same time! For example: If H says unstuck, N and therefore X would get stuck and H would be wrong. If H says stuck, N and therefore X will be unstuck. N is a problem that H can't solve unless H gets more complicated. Its a paradox that's really easy to understand if you pay attention.
@fasty9310 күн бұрын
For those who don't understand, here's what I've observed: H knows how the machine works, so, he will give the answer he believes is correct. So, he will say "stuck" if unstuck and vice versa because he knows N will invert the answer. This is where it stumped me, because since H knows how the machine works, can't he just give the opposite answer then?
@chaumas10 күн бұрын
If H gives the opposite answer, then that answer will also be wrong. No matter what, X will do the opposite of what H says, so H can’t get the answer right.
@fasty939 күн бұрын
Omg no way its chaumus I heard a lot about you in the comment section
@FrankAnzalone11 күн бұрын
Anyone have a practical application for this
@jacobromeo189011 күн бұрын
Do you have the program file you used to make this? WOuld be very helpful when trying to solve problems in my Classsical Mechanics course, thank you!
@mee_is_sus12 күн бұрын
If you give X it's own blueprint, there would be no correct output. So, H would get stuck, and send nothing to N, causing X to output nothing, which IS the correct answer.
@chaumas12 күн бұрын
There is never "no correct output" for H. Machines either run forever or they don't. There is no third option.
@mee_is_sus12 күн бұрын
@chaumas H will receive the blueprint for X and simulate it, then the simulation simulates it and the simulation of the simulation will simulate it, then the simulation of the simulation of the simulation simulates it, and so on forever, causing H, to take an infinite amount of time (and RAM space) to process it. And since the only correct answer to "Will the next word you say be no?" is never saying anything ever again, X's output would technically be correct.
@mee_is_sus12 күн бұрын
@chaumas I guess you're right, there is no option that is both 0 and 1... Wait... QUANTUM COMPUTERS HAVE QBITS THAT CAN BE BOTH 0 AND 1 AT THE SAME TIME! H WAS A QUANTUM COMPUTER ALL ALONG!
@mee_is_sus12 күн бұрын
What if H gets stuck instead of N?
@chaumas12 күн бұрын
Then it wouldn't fit the definition of H.
@qewqeqeqwew397712 күн бұрын
Actually a useful explaination.
@anamberangel13 күн бұрын
Amazing work!
@anushkachouhan599613 күн бұрын
ohh man 17 years ago
@notaspy913 күн бұрын
I don't understand why people in the comments section act like Turing is a fraudster and the Halting Problem is wrong. If you can prove that Turing was wrong, you should submit your proof to a journal! You'll become a multi-millionaire overnight and remembered for hundreds of years as the person who cracked open the fundamental limits of computers! Unless, of course, you're wrong, and the people who make a career out of analyzing this stuff know more about it than you.
@chaumas12 күн бұрын
The only quibble I have with your comment is the claim that you’d become rich. Our society doesn’t value academic contributions that much. You’d for sure get your name in a textbook though.
@notaspy912 күн бұрын
@@chaumas I was thinking about that. Maybe a better strategy would be to show it to a private company under an NDA and agree to license it to them for a royalty? I imagine Google, Amazon, etc. would be willing to pay any amount of money for something like this, but also I don't know how copyrightable/trademarkable a proof is. Or maybe you could figure out P = NP from this somehow and claim the Millenium Prize.
@brianlittle71713 күн бұрын
This is way above my head
@brianlittle71713 күн бұрын
We don’t have tides on the moon. That doesn’t seem fair.
@wavydaveyparker10 күн бұрын
We don't have an ocean on the moon. Who said life was fair.
@brianlittle71710 күн бұрын
@@wavydaveyparker lol
@brianlittle71713 күн бұрын
What is a fictitious force? And what would happen if a cop used excessive fictitious force?
@wavydaveyparker10 күн бұрын
It all depends on which reference frame you're using. If a cop used an excessive fictitious force, then that would imply that you were accelerating, and they could claim that they were at rest, and you put the cuffs on yourself.
@brianlittle71710 күн бұрын
@@wavydaveyparker they could claim I committed a fictitious crime.
@wavydaveyparker10 күн бұрын
That all depends on which reference court the case hearing was being held. In the inertial court, there was no crime - And you'd get off scot-free! If however, we were in the non-inertial courtroom - then your honour, 👩⚖️ the defendant is guilty of fictitious charges. And you're going down, I'm sorry too say.
@brianlittle71710 күн бұрын
@ that’s funny
@nhungduong323613 күн бұрын
Is M a multipurpose computer with 3 inputs and 3 outputs?
@Mystical-honk13 күн бұрын
You are going to LOVE LLMs
@notaspy913 күн бұрын
LLMs can't solve the Halting Problem either.
@omarfabbri577417 күн бұрын
it's crazy how it's simple to understand tensors after this video, Nice work
@fgdgtryhdfgrsgrtsr174919 күн бұрын
Therefore we need a third output: Fatal exception: Russell's paradox occurred Just like accidentally dividing by zero
@julianw101118 күн бұрын
In this specific case we want to show that, when just limiting to "Halts" or "Does not halt", the machine cannot give a logically plausible answer. Which does not make sense, as intuitively every machine either holds or not. There is no "inbetween" super position of halting and not halting at the same time. A machine either halts, or it doesn't. The interesting result is: You can define for every machine M a set of inputs T for which M halts, and a set of inputs ⊥ for which M does not halt. But: It is impossible to create a single machine H which correctly evalutes for an arbitrary machine M and input I whether M on I halts, or not. And there is no inbetween. The example in the video shows, though, that it is impossible to say for a machine with logical consistency whether this specific input halts. It shows that Computers are *unable* to say with correctness whether an arbitrary program halts or not, because Computers are logically unable to do so
@chaumas17 күн бұрын
Can you define precisely when your version of H gives that “exception” output?
@Spudd5019 күн бұрын
These videos can't come fast enough, they are so quality
@BWNAO-d9k19 күн бұрын
Bogo sort? more like bogus sort.
@darrenlo980220 күн бұрын
H can’t be built, but if we modify it to get stuck when it can’t have the right answer, wouldn’t it work in some way now? Sure, it can’t solve all of halting problem, but the new machine might know what to do when H doesn’t. Yes. I know H doesn’t exist. This is just a suggestion.
@whitemouse246019 күн бұрын
How would you know if H is permanently stuck or just needs some more, finite, time to give the answer?
@chaumas19 күн бұрын
The problem is that “when it can’t have the right answer” isn’t possible for a machine to determine. Let’s be specific: we have a machine we’ll call “Almost H”, or AH which often answers the halting problem correctly, but sometimes gets it wrong. We want to build a second machine we’ll call “Check Almost H” or CAH which returns true if AH would give the right answer, and false otherwise. Well, here’s the issue. I can build this machine now 1. Pass blueprint and input to AH to get “initial answer” 2. Pass blueprint and input to CAH to check AH 3. If CAH said true, return the initial answer 4. Otherwise return the opposite of the initial answer And oops, I just built H, which is impossible. So CAH must also be impossible. The upshot is that yes, you can build partial halting solvers, but they must either give the wrong answer in an infinite number of cases (and we can’t tell which they are), or it must give a non-answer in an infinite number of cases, some of which are halting cases and some of which are non-halting cases. A key point here is that there is no such thing as an individual machine and input whose halting answer is undecidable. Either a machine that always says “stuck” or a machine that always says “not stuck” will get the answer right for a given case. It’s when we try to build a single machine that gets it right in _every_ case that we hit a contradiction.
@Rpem1020 күн бұрын
Now I need to know what if H negated it's own output id it was fed with 2 X blueprints.
@goldfishglory20 күн бұрын
solution: solve every problem by creating a machine to solve that problem. if the machine is like x and contradicts itself, then don't make it. if you decide to keep it, make a re-routing machine. but then, if you want it to do everything, you'd need a categorizing machine. but what if you feed the categorizing machine something it can't handle? make more categorizing machines. eventually, there will be too many machines. solution, make one final machine that rejects an input if it detects a problem. i imagine this to be a large machine line. you have Z, the categorizing solver machine that is a mix of a ton of computers, and then you have I, the imagine machine that comes up with problems, and then O, the output machine that feeds the output back into I.
@paulblart737820 күн бұрын
So... your solution is to ignore the problem?
@goldfishglory20 күн бұрын
@@paulblart7378 yeah! just use machines in the way they're intended. they may not be able to do *everything*, but if you're gonna use them don't torture them ._.
@paulblart737820 күн бұрын
@@goldfishglory ...I guess that makes sense. That has no relevance to this proof though.
@darrenlo980222 күн бұрын
7:23 You can prove that a few programs do not halt. The program shown keeps running the loop and has no condition to end the loop.
@genericobjectshow344022 күн бұрын
... so H is just a hardware emulator. Like a hyper-advanced CAD software.
@chaumas22 күн бұрын
No, H is not an emulator. An emulator takes in a description of a machine and an input and does whatever the machine would have done with the input. If the described machine doesn't halt, the emulator doesn't halt. If the emulator halts, you know that the machine described will halt. If the emulator _doesn't_ halt, you can't tell the difference between "hasn't halted yet" and "won't ever halt". H is different. It takes a description of a machine and an input, and _tells you_ whether the machine described will ever halt. This is a completely different task from emulation. For example, take this machine: 1. Take in a number as an input 2. If the input is even, print out a billion smiley faces and then halt 3. Otherwise, print out a billion frowny faces and go to step 2 Now, does this machine halt if we feed it "7" as an input? No. In this specific case, we can see that instantly, and we didn't have to emulate anything. We can figure out the answer without writing down a single frowny face, much less a billion. But if you feed that description to an emulator and give it the number 7, the emulator will never tell us whether the program halts. It will just keep printing out frowny faces forever.
@genericobjectshow344021 күн бұрын
@@chaumas ...fair enough.
@genericobjectshow344022 күн бұрын
I find it funny... In the intro, A and C are introduced... when both are just A. C is just A with different instructions.
@Lolek14-vk8rf23 күн бұрын
honestly this is like disproving calculators cause they cant divide by 0, give me more of this content
@chaumas22 күн бұрын
No, it's not like that. Dividing by zero is not defined. A calculator that prints out "undefined" or "error" when you try to divide by zero is not a contradiction; it is behaving correctly according to its definition. The question of whether a machine halts on a given input _is_ always defined. It always has a single, concrete answer. But when we imagine a machine capable of consistently _finding_ that answer, we see a contradiction.
@Lolek14-vk8rf22 күн бұрын
lemme edit the comment so its more defined
@imagiro123 күн бұрын
And did I just understand what the Quantum field theory is about? I start to think this might be one of the best videos I ever saw on YT. Next to "Which way is down", which let me understand General Relativity.
@imagiro123 күн бұрын
Did I just understand why the sun is always there when I look at it, but a photon from the sun may or may be not?
@lawhorn26 күн бұрын
Pigeonhole sort is just average human organizing
@drgeetchadha26 күн бұрын
Everyone's sigma till they bring out quantum tunneling😅
@BarbaraTaylor-c1u28 күн бұрын
The video is very interesting! Something I don't understand: I have USDT in my OKX wallet and I have the recovery phrase. pride pole obtain together second when future mask review nature potato bulb: How should I convert them into Bitcoin?
@geeteevee766728 күн бұрын
3:01 i find it so stupid that the metal bar man puts the same machine blueprint in twice as if it isn’t already in there
@flash165227 күн бұрын
That function (or in this case, machine) needs 2 inputs. You can't say "use the input that you already have", usually.
@wintutorials2282Ай бұрын
I’m watching this with my autistic girlfriend INSERTION SORT WAS LACKING BRUH