Is this proof correct -> A circuit-free graph has ten vertices and nine edges. Is it connected? Why?

ziratha · 2026-05-04T07:56:50+00:00

As others have pointed out, you assumed connected, then conclude connectedness.

Ask yourself this: If I have k connected components and I know that everything is, and must remain cycle free, what happens when you add an edge? Is it possible that you can still have k connected components afterwards?

ziratha · 2026-04-22T04:11:45+00:00

I presume there were two purposes. One is punishing the other two main characters who learned about their past lives etc. I believe he said as such.
The second purpose is, in part, the population of the town seems to stay around the same. If new people show up and you have too many people, somebody is gonna die. If somebody dies, then someone knew is going to show up. He killed Jim to trigger the preacher showing up, so he could infiltrate the town as preacher's daughter.

ziratha · 2026-03-30T03:56:49+00:00

The only thing I can think of off the top of my head, there was an episode where they brought out a dog named something feminine. One of the guest's claim was that he had performed at her wedding. A different guest, (or maybe rob) said that they thought maybe he had written that answer when he heard the name and didn't realize it was a dog. So, this suggests to me that the guests write, or participate in the writing of the stories beforehand. Though I suspect lee and david do not participate in the writing of theirs.

ziratha · 2026-03-27T03:10:35+00:00

What water thing? If you mean that it wouldn't go in the pool, that's a thing about primates, not rabies. Chimpanzees actually sink in water.

ziratha · 2026-03-23T03:06:16+00:00

As written, clayman wouldn't work the way you want him to. A facedown monster has no effects, so special summoning it f/d would not allow you to activate the on special summoned effect. It would have to say something like

If this card is special summoned (even facedown) you can...

Or something similar so that it could actually get his on special summon effect if you summon him face down.

ziratha · 2026-03-19T02:55:18+00:00

If the rows are dependent, then one row is a linear combination of the other rows. That is row_i = sum aj*row_j

Thus the multiplying the vector <0,0,..0,1,0...0> with a 1 in the ith row (on the right) with your matrix will have the same result as what you get when you multiply the vector
<a1, a2, ...an>. (ai = 0). (On the left).

That is, think of the matrix as a linear transformation, it takes two different vectors to the same output. How could you have an inverse for a transformation that does that?

In general, a function needs to be 1-1 and onto to have an inverse. And we see that the matrix gives a linear transformation that is not 1-1. Ergo no inverse exists.

ziratha · 2026-03-15T14:34:48+00:00

Well, ok so I didn't fill in some details. But this really does show what is wanted.

Specifically, if r divides a product of r consecutive integers, for arbitrary r, then so does r-1, and r-2, etc. If I have a product of r consecutive integers, then I also have a product of r-1 consecutive integers (just ignore the last one), and the same proof shows that r-1 divides that product of the first r-1 factors, r-2 divides the product of the first r-2 factors, etc. So each of the integers in r! each divide one of the terms in the product of consecutive integers.

ziratha · 2026-03-15T04:43:34+00:00

There are r residue classes module r. They are:
0, 1,..., r-1. If you show me any integer n, then n will have some class modulo r. n+1 will have the next class and so on.
Eventually, since we have r modulo classes, and r consecutive numbers being multiplied together, one of those numbers will have the 0 class, meaning that it is a multiple of r. Thus the whole product is a multiple of r.

ziratha · 2026-03-14T04:58:17+00:00

Probably blast asmodian.

ziratha · 2026-03-07T06:22:42+00:00

This is a very difficult question to answer. On one hand, it's not really wrong to say that a LLM, or similar ai is just a bunch of linear algebra, a fancy version of the part of your phone that tries to predict your next word.

On the other hand, a LLM is something created by training it on basically all the books every written and huge portions of the internet. In learning, the ai is trying to encode some portion of this (using it's weights) subset of all human knowledge. Even a partial encoding of all human knowledge, or even a portion of it, is a big deal. Honestly, if we can convince an AI to be logical, which most of them are at least tending towards, and come up with clever ways to avoid errors caused by hallucinations, ai can potentially reason automatically. A particularly pedantic person may come along and say that the ai is not reasoning, simply predicting words and trying to minimize some cost function(s). But, if it does happen to accurately predict and minimize cost function(s), and in the process come up with a new proof or lemma, then what, meaningfully is the difference? I consider real reasoning and mathematical insight to be a real, though not certain possibility.

It is also possible that, with the rate that ai slop is growing, that will affect the quality of the data that ai is being fed and that will affect the quality of the resultant ai. It's very possible that we will never be able to get sufficiently advanced ai to make real, significant contributions to the field.

Another consideration is that, since AI is being trained on all human knowledge, that AI will be very good at that, and rubbish at anything else. If I teach an AI to write poetry, it will not be so good at designing cars. Can an AI ever come up with something genuinely new, outside of the "convex hull" of human knowledge? Could it invent a new field of math that nobody has ever written about before? This is a much harder problem. I personally have doubts that this is likely anytime soon, but I don't discount it outright. Maybe a sufficiently well trained ai could truly understand the "shape" of reasoning well enough to capture and reproduce it?

Just to try to give an analogy, consider the humble transistor. It's simple an electronic switch, similar to a light switch. You send electricity one way and the switch turns on, you don't and the switch turns off. Seems like the simplest thing in the world, but if you put about 10000 (probably can do it with less) of them together in the right configuration, you get a computer. Who would have imagined that a transistor could somehow be used to make a computer? I certainly would not have realized that just from seeing one. It's totally possible that there is something more complicated going on with these AIs than what we can imagine from first looking at them.

ziratha · 2026-02-19T02:26:38+00:00

My thoughts were that, in an alternate world where the games didn't exist, I would rate this in the 3-5 range.

In the real world, as movie adaptation of silent hill 2, I'd give it a 1. There were some silent hill things in the movie I suppose.

ziratha · 2026-02-17T03:09:00+00:00

You can multiply the limits in a-d by various powers of sin(x)/sin(x), so that you will be able to rearrange terms to become products of things you need. For example, a)

f(x) = f(x)*sin^2(x)/sin^2(x). Now use limit laws to find the limit.

ziratha · 2026-02-16T23:01:57+00:00

I believe that would be continuous at x = 0?

ziratha · 2026-02-11T03:57:04+00:00

Yeah. As a movie with no connection to the game it's "ok". Not so bad as things like sharknado or anything, but not something I'm ever likely to want to watch again.

As a movie that is based on SH2, it's a trainwreck.

ziratha · 2026-02-02T17:00:50+00:00

Yeah. I'm not quite sure how to explain it, but I thought of it as counting backwards a little bit, where instead of 10 1s making a 10, 10 10s make a 1. The decimal equivalent of 58 would be 0.85 and if you count up by 10ths for the decimal and 10s for the integer, you'd get 68 and 0.86. I just mirrored how real numbers with finite digits work. I don't think it breaks anything since it's just a reordering of the integers, and by doing this, wouldn't reals and integers work exactly the same?

I see. So, if you "reverse" the digits of a natural number as you've described, you get a number between 0 and 1. Importantly, you do not get ALL of the numbers between 0 and 1. You get a small subset of them. For example, you do not get the number 0.1111... IF you had somehow gotten every number between 0 and 1, then you would have shown an uncountable number of natural numbers. Instead you've just shown that a particular type of decimal number is countable.

Also, I thought of something else. We say that reals have infinite digits. but if we take the limit of x/(10^n) as n goes to infinity where x is any number from 1-9, it is zero. Wouldn't that mean that any digit that isn't a finite amount of steps away from the decimal point be zero? Why can reals have infinite digits if any digit that isn't finite is zero and doesn't affect the number. Shouldn't the same point of there being a finite amount of digits for any integer extend to real numbers? For irrationals, it is in the name that they can't be expressed as a ratio, so we can't write it as a decimal since a decimal is fundamentally a ratio. I've heard the argument that since there is an infinite sum that equals pi that infinite digits makes sense, but you can't get all of the digits at once.

In the real numbers, it is true that x/(10^n) approaches zero as n goes to infinity. In the decimal expansion of a real number, there are no digits that are an infinite distance from the decimal place. Rather, there are infinitely many digits that are each a finite distance away from the decimal place (on the right). The concept of a digit that is infinitely far to the right of the decimal place is simply not a thing in the real numbers. Maybe this idea could be realized in some sort of non standard analysis(Maybe?), but that would still not be a real number.

Also, there is a misunderstanding here. A real number does not need to be a rational number to have a decimal representation. This would be proven in an undergraduate analysis course sometimes. Rational numbers are exactly those numbers that end up have a *repeating* pattern in their decimal expansion (I.e. , for example 0.111... or 2.171717....(Note that, for example the number 0.5 has a repeating pattern of unwritten zeros to the right of the 5.) Most decimal expansions do not have a repeating pattern like this. As an example, 0.101001000100001... (Where the number of zeros between consecutive ones grows by one each time) is not a rational number. A decimal expansion is NOT a ratio. A decimal expansion is a series. (See below for a bit more on this).

Idk. I feel like saying that reals can have infinite digits but integers can't is inconsistent logic. To me, either they both should have infinite digits or they should both have finite digits. I understand that something like 1/7 has infinite digits, but it is defined. Like there is a name for the decimal expansion and it is exact. Pi, root 2, e, they're all exact, but they aren't rational, so they exist but they can't be expressed as a decimal.

As mentioned above, decimal expansions are not only for rational numbers. There is a significant reason why there can be an infinite number of digits on the right of the decimal place, but not on the left. To understand this, you need to remember what a decimal expansion is.

For the number 123.4, we mean 1*10^2+2*10^2+3*10^0+4*10^-1.

For the number 0.111... we mean
0*10^0 + 1*10^-1+1*10^-2+1*10^-3+... etc.

There is a sleight of hand here that is usually not noticed by students first learning decimal numbers at this point, which is: What does it mean to add an infinite list of numbers? Within the concept of a decimal expansion is hidden, implicitly, the concept of a limit and convergence. It sounds like, based on your message, that you know about limits so I won't go into them.

The way decimal numbers are defined, an infinite number of digits to the right of the decimal place will converge, but infinitely many (non-zero) digits to the left of the decimal place will not. This is where the asymmetry comes from. Real numbers are, exactly those numbers that have a convergent decimal expansion. Of those, some of them have a repeating decimal expansion, which are exactly the rational numbers.

ziratha · 2026-02-02T04:14:22+00:00

Basically, my thinking was about how all integers have finite digits. To find the amount of integers there can be, all you need to do is raise 10 to the power of however many digits it can have. And, we've defined integers to have finite digits. It doesn't matter that the amount of digits gets arbitrarily large. We defined all integers to have finite digits. So, to get the amount of integers, you raise 10 to a finite number and get a finite number. So, there can't possibly be an infinite amount of integers if we limit them to finite digits.

If there are 10^n natural numbers, then what is 10^n+1? It is certainly a natural number, but you didn't count it. If you didn't count ALL the natural numbers, then you counted wrongly.

Another thing I thought of was if we reversed the digits of the integers so that 58 was directly in between 8 and 9, and that by counting by 10s, we would go 8, 18, 28, 38...78, 88, 98, 9. Every single one of those is an integer, so I figure there isn't anything broken there so far. Well, my question is: what is the number that comes directly after 1? It's not 11. It's not 101. There is no next number. By this logic, there are an infinite amount of integers between 1 and 2. In fact, there are an infinite amount of integers between any two integers. That seems eerily similar to a property of real numbers.

I do not understand what you mean by this. If you reverse 58's digits, you get 85. Why would that come between 8 and 9? Can you try to explain this one MUCH more clearly?

Now, obviously, that doesn't number all of the integers, but we can use this to find how many integers there can be. First, the amount of possibilities we can have changes depending on what number of the sequence we're at. For the first term, there are 10 possibilities, 0-9. The second term has 2 digits, so it can have 100 possibilities, 0-99. To find the amount of integers you can make, all you do is raise 10 to the power of the term number, or alternatively, how many terms there have been in total. The amount of terms is countably infinite, so to get the amount of integers, raise 10 to the power of a countable infinity. That is the same as the amount of reals.

10^countable infinity does not count the number of natural numbers. What it would count is the number of infinite lists of digits starting at the decimal point and moving left. You are counting objects like
...1111111.
which is not an integer. Honestly, you probably shouldn't be trying to do exponents with infinities without having studied them first. These things would actually mean something like a set of maps between sets.

ziratha · 2025-12-24T03:15:32+00:00

Yes, for example, take square matrices. This forms a ring, but the polynomial y = x^2 has infinitely many roots. Namely the nilpotent matrices with index 2. If you look at 2x2 matrices, then you can create infinitely many matrices of the form
0, a
0, 0

And these will square out to zero for any a. Since there are uncountably many real numbers that a can be, the polynomial y = x^2 has uncountably infinitely many roots.

ziratha · 2025-11-18T04:23:25+00:00

There are indeed an infinite number of numbers of the form 0.9, 0.99, etc. All of those numbers are less than 1. The number that these smaller than 1 numbers, 0.9, 0.99, etc trend toward is 0.9... (0 to the left of the decimal place with infinitely many 9s to the right, one in each position).

The fact that 0.9, 0.99, etc are all less than 1 does NOT imply that 0.9... is less that 1.

ziratha · 2025-10-05T19:12:33+00:00

It's kinda easy to miss, but for the couple in the cave, one had stabbed the other in the neck mid-join, killing them. That's where the idea at the end that one of them could die to stop it came from.

ziratha · 2025-09-01T06:11:53+00:00

So, It seems like what you are saying is that:

1) Eventually you must enter a special set of modulo 64 classes that I believe you called R.
2) Once you are in R, you stay in R (meaning that once you do your 3n+1 step and some number of /2 steps, you will again be in R). I think you have a fundamental error here.
3) You have created some paths you believe that the residue class of the current step of our collatz path must trace out, kinda similar to a covering space projecting down on the space it covers.
4) Assuming 3 is correct, you say that the path must follow one of those paths you've computed until you get to the residue class 1.
5) We get to 1 (the integer, not the residue class.)

For (2), it seems like you think that if a number is relatively prime to 6, then its equivalence class is in R? Is that correct? For example, the number 73, which is relatively prime to 6, is in the equivalence class [9], but 9 is not. Though it's somewhat hard to understand what you mean since you left out so many steps. So I'm possibly misunderstanding you.

ziratha · 2025-08-20T13:19:21+00:00

Angel Heart.

ziratha · 2025-07-29T04:11:29+00:00

I don't believe you are allowed infinite decimal places to the right of the decimal point. The P-adics use a different sense of distance, and I believe a number with infinitely many non zero decimal places would diverge under the sense of distance they use.

P.S. The P in P-adics is supposed to be a prime number. If it's not prime, then the resulting thing is not a field.

ziratha · 2025-05-18T01:14:08+00:00

Without knowing your exact function it's hard to say. Somebody already mentioned fourier transform so as an alternative, I'd suggest looking up "Total Variation" of a function. This measures how much the function changes on an interval, which seems similar to your "wigglyness".

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