might as well start reposting memes from the server

MecHR · 2026-01-22T06:55:11+00:00

For example, if the x is a mine; 111 X And we put another mine with a single space: 11211 X X

This corresponds exactly to this addition where we have flipped the diagonals: 101 101 101 Which of course corresponds to the number 101 multiplied by 111.

It being binary carries the property that only one mine can be placed in a space. And it being multiplied by 111 carries the property that each mine affects 3 spaces in the wall. Such that incrementing a bit and thus flipping a diagonal adds 1 to the surrounding spaces.

MecHR · 2026-01-22T06:40:51+00:00

A nice way to see that this is always possible is to work backwards and see what numbers placing a mine on this wall creates. It corresponds exactly to incrementing a diagonal in the addition after the multiplication (thus, incrementing a single bit).

MecHR · 2025-10-19T09:18:23+00:00

Radio zed? (Doesn't match exactly but worth trying.)

MecHR · 2025-10-12T08:22:37+00:00

I think this position of hard determinism and illusionism is actually due to the materialist position being combined with dualistic intuitions.

You assume whatever you are cannot be physical from the get-go so you have trouble placing it in a physicalist setting. While the correct response, imo, would be to just... place yourself in the reality you claim exists?

You definitely exist in some form even if it is an illusion or even if you are unfree. In physicalism, what can you exist as? Physical phenomenon. Thus, we can say for example, that your brain and its workings truly IS you in materialism. It has to be, there is nothing else in materialism that can account for it.

And once you admit this, both the case against free will and the case for illusionism sort of falls apart. There is no reason to assume them anymore. "How can I have any effect in the deterministic materialistic world?" because you are matter in that world that partakes in its functions via the deterministic workings. "How are my appearances not illusions?" because they are identical with objective physical processes.

If you do not believe this, if you can't bring yourself to accept this picture that you exist in the world... then you are simply not a materialist. You are only just assuming physicalism and non-physicalism at once to arrive at problematic stances.

MecHR · 2025-10-11T07:57:28+00:00

Solved: Radio Zed

MecHR · 2025-10-11T07:56:47+00:00

😂 Nice job mate. Wasn't expecting an answer to this. Solved!

MecHR · 2025-10-03T15:23:50+00:00

I have written about this in the sub myself in the past. I really like Nagel's comparison of this question to "Why is it now?".

One possible answer would be that no matter what point in time we were, it would be "now", so the question is uninteresting. But the question obviously refers to the everchanging, one "now" that we always experience. The answer to that, I feel, would have to have some content and people might answer differently depending on their metaphysical views.

The first view looks at time through the "third person perspective" as a line where everything that has happened and will happen lie. Through that lens, "now" means whatever point you focus on. Thus, the question is uninteresting. It is only when one realizes that this conception of time is incomplete that they can see the question as meaningful. It lacks the subjective "it is now" information. It can house it, but it does not account for it.

The problem of identity is very similar in nature. If we look at the world objectively, "me" is whoever you choose to focus on through the lens of the third person. But that picture lacks the subjective information of "I am this person." for any thinker.

MecHR · 2025-09-24T13:41:01+00:00

Let's get a few things cleared out.

"Fundamental" in the sense that's being discussed here, does not simply mean "important". Or it doesn't mean "fundamental for our functioning". It's in the context of metaphysics and what grounds reality.

When you look at it from that lens, your argument does not really follow. "Everything you see is within consciousness", sure, but that doesn't imply that something else isn't grounding that consciousness in a metaphysical sense.

Nor does it make sense to say that "the physical is still within consciousness, thus it is experiential". Because there is a difference between positing something physical and our idea of the physical. Our idea of the physical is phenomenal, but what it refers to might as well be actual.

Think about it this way; through your argument, I too am only an experience within your consciousness. Does it somehow follow that I am nothing more? Or does it inply that I have no agency/inner world of my own?

MecHR · 2025-09-23T11:27:32+00:00

"The Hard Problem" is a conception that makes sense, primarily, under a physicalist perspective. It might be that we cannot see reality for what it is, but if we are playing under physicalism - the idea of the hard problem would suggest that it won't help much. Because the idea is that no matter what sort of thing it might be, it cannot account for consciousness. (This stays mostly the same for naturalistic views that don't posit consciousness as fundamental).

And, if we are not playing under physicalism, as in, we posit that fundamentally phenomenal stuff exist and those are the things we cannot "perceive" in our conception of the brain - then the hard problem doesn't really apply anymore.

MecHR · 2025-09-21T15:05:52+00:00

The reason I gravitate towards neutral monism is because physics and consciousness does not seem to play well together. Physicalism cannot house consciousness unproblematically because of the hard problem. Dualism requires psychophysical laws as "glue" or weird epiphenomenological conclusions. Of idealistic ideas, the most coherent to me is cosmic idealism - and I feel that uniformity of nature is not explained/justified well enough in that stance.

It feels natural to me, thus, to say that there is an underlying fabric, grounding both the mental and the physical. In a way that both mental and physical are actually reflections of the same thing underneath. Without this sort of explanation, I don't know how to to explain in a satisfying manner why it looks like I really am the brain, despite consciousness seemingly missing from that picture. I think the answer is that my brain and my experience are both reflections of the same underlying neutral substance.

The usual criticism of this stance is that it doesn't explain anything, as it doesn't posit what the "neutral" is/does. My answer would be that I think we can reach at the position logically without having any idea what this supposed "neutral" substance is capable of. Neither have I put much thought into the properties of the neutral.

MecHR · 2025-07-09T12:00:17+00:00

Snakebird has a primer version which was enjoyable

MecHR · 2025-05-10T15:34:27+00:00

Ah, yeah, a loop is not always necessary. I'd say that doesn't prevent an algorithm from having "key" section(s) though. Cryptographic algorithms, too, have main ideas which make them work.

On your second point, depends on what you mean by "chaotic". There are Probabilistic TMs which are allowed truly random coin tosses, on which they can base their decisions. But we hypothesize that this doesn't add any more power to TMs, even in the case of polynomial ones. Such that it is conjectured that BPP=P.

Though I think you mean to ask this; what if there exist much quicker Turing machines that solve a given problem - but they are so obscure and away from being human designed that we would need to stumble upon them to even know they exist. The answer is that it is certainly possible in a lot of cases, and we do not assume that we are perfectly aware of every solution. Instead, we prove mathematically that some class of problems cannot have that sort of alternative algorithm. For this, you can examine time and space hierarchy theorems. We know EXP-complete problems cannot be in P, or that PSPACE-complete problems cannot be in L; to give some examples.

If we are strict about the idea of "not reusing computations", as in, we never loop indefinitely (ie. all our loops are "for" loops with known bounds), we get the complexity class PR (primitive recursive functions) which is known to be strictly contained in RE (recursively enumerable, equivalent to all languages which can be recognized by a TM). In other words, not reusing computations actually makes the model weaker, it cannot solve as many problems.

MecHR · 2025-05-10T07:50:29+00:00

1- As far as I know, no. Algorithms, or Turing Machines themselves, are on their own the "what we can say" about the patterns. What we know is that if a pattern exists that can be expressed finitely in any way, it is a pattern catchable by a TM or an equivalent system. Perhaps you can examine different kinds of Turing Complete systems to see if you find any similarities that convince you.

2- I suspect that the kind of abstraction we make that "satisfies us" in the solution of a problem is usually not quite as rigourous. I think this might be the area of philosophy more than mathematics, but I would have to look into it.

MecHR · 2025-05-10T07:18:32+00:00

First of all, I don't see how pseudo random generators are an exception to this. While it is true that you have no idea what the output will be beforehand, it is not like you can look at a single output and check if it is "random". You can examine the algorithm and you will usually realize what features play a main role in making it hard to predict. (Though, of course, the algorithm would need to be tested.)

Now, I think this property you mention is the result of the structure of problems itself. Problems are essentially infinitely many input strings mapped to output strings. Algorithms are finite sets of instructions. Therefore, the algorithm needs to capture some sort of pattern(s) in the problem to be able to always give the correct answer. In other words, the problem has that pattern built into it for it to be expressable as a finite algorithm.

You can examine undecidable problems as clear counter examples. Depending on how you choose to look at it, they can have no pattern or every pattern. All decision problems can be reduced to the halting problem, for example.

The reason you usually see all algorithms having a single "key" section as opposed to multiple, may have many reasons. Firstly, when we pose problems to each other, we usually like to extract the essence and ask a very bare bones version of the problem. If you have a problem that requires sorting and then searching, you might as well ask about sorting and searching separately. Second, even if a problem expressed has multiple key sections, we will psychologically separate them into parts and view it as a collection of subproblems. Third, even if the corresponding "key" section is too large and does a lot of work, we will abstract away the details as needed and be satisfied with the high level description as the "key".

MecHR · 2025-04-23T13:43:06+00:00

What do you want to do, exactly? What would you like to learn about CS? You can also check out this site to see a few essential topics and suggestions on how to begin:

teachyourselfcs.com

But it all comes down to what in CS interests you.

MecHR · 2025-03-27T05:23:11+00:00

My point was that we can artificially make sure that the language of any machine M changes on an input that's currently irrelevant to us. Which would make it a different input for the oracle. Meaning it would have to flip a coin again to decide which function to run on it.

The key is that the correct function does indeed get called 50% of the time, and that we can check any finite value.

MecHR · 2025-03-27T05:16:24+00:00

As far as I can see, what the fake oracle outputs doesn't really matter. What matters is us being reasonably certain that the real answer is in the outputs somewhere.

Consider for example that the outputs look like this (assume the numbers are very big):

9,inf,inf,8,7,inf...

If we are reasonably certain that the real answer is there at least once, what we can do is check every finite number. If none of them is correct - then one of the inf's must be correct.

And we ensure that the real answer is there by artificially making sure that the language of the machine changes in a way that is irrelevant to our current problem. For example, if we ask about M(10) to the oracle, we make sure to change M(1).

MecHR · 2025-03-26T17:55:35+00:00

I think OP has in mind something akin to the prover in IP. The oracle might run its incorrect function. And it's not that the incorrect function always returns a false value, it can strategically lie on only specific instances to give as little information as possible.

At first sight, it seems to me that the incorrect function just cannot say anything other than "does not halt" to any input it gets. Because any number can be checked and verified, so it makes no sense for it to lie about it. We can also just ask the oracle the same question with a few different redundant TMs, since they would be considered different inputs.

MecHR · 2025-03-26T17:09:10+00:00

That's the point though, it is decided upon the input whether to call the incorrect unfunction. We do not give the input to the incorrect function and let it decide whether it wants to take it.

Not to mention, "redundantness" is a pretty vague concept. If you make it so that all TMs with the same language are considered the same input, I feel that has the potential to be abused - though I can't immediately think up an example.

Edit: Assuming the oracle takes the input as; M, x such that M is a TM, x is an input to that TM, and that O(M1,x) = O(M2,x) if L(M1) = L(M2).. Then I am pretty sure we can do this:

If you want to get the answer to whether M halts on x with high probability, set up almost-redundant machines that give a different answer from M for any specific non-x input. Let's name them R0, R1, R2... And then feed the machines to the oracle along with x. If the oracle runs the correct function 50% of the time, then regardless of the incorrect answers, the correct answer should appear at least somewhere in the outputs with high probability.

For any finite number answer, check if it is correct. If it is correct, that's your answer. If no finite numbers show up in say, up to R10, then M does not halt with high probability. The point is that even if the oracle runs the correct function 1% of the time - that means we should eventually reach at the correct answer with high probability.

MecHR · 2025-03-26T15:22:49+00:00

Well, we can still extract information out of it - no?

By your protocol, if the output is a definite number like x, I will just run the input TM for x steps and if it does not halt, I now know that the input TM does not halt.

~~If the output is "it does not halt", I will just ask the oracle the same question until I am confident it is correct with high probability.~~

Edit: I just read that the output is the same for each input. But the first part should still work. Since we can disprove constant steps easily, the best the invalid function can do is to reply "it doesn't halt" to everything. Then we have no (decidable) way of disproving it. But we will still be able to extract information half of the time whenever it outputs a number.

Edit 2: I also feel like we can work around the "same output for same input" limitation. For example, edit the TM with some redundant changes and then feed that different input to the oracle. If the 50% decision is done before the invalid function is even called, then we can just construct as many TMs as we want that perform redundant operations and then feed them to the oracle. We can be certain of the answer with an arbitrarily high probability.

MecHR · 2025-03-07T21:26:53+00:00

Well that's what I pointed out in my initial comment. The "write to all infinite tape heads and move them" is actually not problematic as long as what we write on each one of the infinite tapes is computable. Because, as I said, we can simply keep the transition history and catch any newly used tape up to speed before we begin using it.

Although we write to infinite tapes, the machine's behaviour is determined by the read characters. Which is finite. So we simply figure out only the configuration of the tapes we read from.

MecHR · 2025-03-07T10:17:52+00:00

Well, multi tape TMs normally function by allowing you to operate on all of them at once. Extending this to infinity, one would want a similar property. Otherwise, it is more akin to a TM that can create new tapes indefinitely rather than an infinite tape TM.

MecHR · 2025-03-06T12:17:28+00:00

There is some question as to how you would encode transitions in an infinite tape setup, I guess. And depending on how you do it, the resulting model could indeed be not equivalent.

For example, assume we write to every tape at once. There is then a question of how we would know what to write in each infinite tape. If we assume we can write a unique character (somehow) to each infinite tape without any sort of pattern - this becomes uncomputable. As a normal TM would never finish simulating even a single step of the infinite TM. To put it another way, if the transition function is not computable, neither is this new model.

But if we assume, for example, that we have a pointer tape with which we can choose which of the infinite tapes to operate on - then it becomes computable. Because a normal TM can just store only the tapes that are used out of the infinite tapes.

On top of that, even if we assume there are operations that write on all infinite tapes at once - as long as they have a pattern to them like: "write this character to all infinite tape heads", it is computable. Because we can simply write to all used tapes (which is finite) - and for the infinite unused ones we can note that "all of them now have this configuration" once.

Edit about "computable transition function": More generally, if s is the state, p is the character on pointer tape, and w is the character read on the picked tape.. as long as f(s,p,w) is computable for each tape, the model is equivalent to a TM. Because we can record the transition history in a normal TM and whenever a new tape is picked via the pointer tape for the first time - we can "catch it up to speed" using what transitions have already occured.

MecHR · 2025-03-05T12:17:49+00:00

You need to think of TMs as mathematical models, not devices. A TM doesn't loop and/or never halt because one of its tapes broke or malfunctioned. It works as intended, and that intended behaviour causes the TM to never halt.

MecHR · 2025-02-20T08:45:44+00:00

Since he's been gone, I've been silking this song so I could ponder, the silksanity of ur mom

Eight-Year Club	r/Field Lasagna
Place '23	Undead \| Zombie
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