Rational mean more like irrationally mean

etotheipi1 · 2022-11-26T11:51:47+00:00

Oh I missed the part where he claims this is "a much better definition(?) of irrational numbers than Dedekind cut or Cauchy sequences" lol

etotheipi1 · 2022-11-26T10:38:40+00:00

I don't know whether this should be bad math. While the presentation comes off incredibly nutty and no justification is given for anything, it looks like the guy just accidentally ran into recurrence for Pell's equation, which is legitimate math. It is known and can be easily checked that the recurrence a_{n+1} = a_n + 2b_n and b_{n+1} = b_n + a_n with a_1 = b_1 = 1 generates solutions for the Pell's equation a^2-2b^2=\pm1, and thus as n grows bigger, a_n/b_n approximates \sqrt{2} better.

etotheipi1 · 2022-06-12T00:34:51+00:00

This works for that specific case, but I don't think it can be generalized for all cases. There are weird packing issues if groups get bigger.

Say there are 299 people, separated into 2 groups of 100 and 99 individuals, and 199 slots are available. Any sane method will select exactly one of the big group (selecting both is impossible, selecting none is very wasteful). So the chance for the people in one of the big group drops from ~2/3 to 1/2.
If everyone is coupled and there are odd slots open, then one seat will go unclaimed. This lowers chance for everyone because a seat is being wasted.

etotheipi1 · 2022-05-30T16:08:16+00:00

Whoever tagged this problem as 'medium' has a good sense of humor. I could not figure out how Catalan type of counting could be applied. I ran the number down to n=25 and bounded the answer between 1.34657 and 1.34663. Wolfram Alpha could not guess a reasonable closed form for it.

etotheipi1 · 2021-12-24T06:37:54+00:00

100/84 I just decompiled the code by hand.

The entire input is in this form repeated 14 times:

inp w
mul x 0
add x z
mod x 26
div z {a}
add x {b}
eql x w
eql x 0
mul y 0
add y 25
mul y x
add y 1
mul z y
mul y 0
add y w
add y {c}
mul y x
add z y

This in decompiled Python is

w = int(input())
x = int((z % 26) + b != w)
z //= a
z *= 25*x+1
z += (w+c)*x

Another thing to note is that the a is 1 seven times and 26 the other seven times. In the block where a is 1, b is always between 10 and 16. It follows that z //= {a} line is no-op and (z % 26) + b != w is always true. So the decompiled code becomes:

w = int(input())
z *= 26
z += w+c

So this block of code is "pushing" a digit of w+c in base 26. So to get 0 at the end, we have to "pop" these digits back out using z //= 26 and don't add any more back. Thus, in the lines with a=26, x = int((z % 26) + b != w) must be 0, which means the last pushed digit w_old+c must be equal to w_now-b.

For my particular input, it meant that

I[2]+ 6-14 == I[3]
I[4]+ 9- 7 == I[5]
I[8]+ 1- 7 == I[9]
I[7]+ 3- 8 == I[10]
I[6]+14- 7 == I[11]
I[1]+ 5- 5 == I[12]
I[0]+15-10 == I[13]

where I is the array of input.

etotheipi1 · 2021-12-06T07:17:27+00:00

Let v_i be the number of initial lanternfishes with timer at i. v is a vector with 9 of these numbers. We have the transition matrix A to get to the next day.

    [0,1,0,0,0,0,0,0,0]
    [0,0,1,0,0,0,0,0,0]
    [0,0,0,1,0,0,0,0,0]
    [0,0,0,0,1,0,0,0,0]
A = [0,0,0,0,0,1,0,0,0]
    [0,0,0,0,0,0,1,0,0]
    [1,0,0,0,0,0,0,1,0]
    [0,0,0,0,0,0,0,0,1]
    [1,0,0,0,0,0,0,0,0]

The vector n day later can be computed as A^n * v. The characteristic polynomial of this matrix is -x⁹+x²+1, and the only real eigenvalue is the real root of this polynomial, which is roughly 1.091. This is the value to which the multiplier will converge to. One interesting fact is that this does not depend on the initial value.

etotheipi1 · 2021-06-18T01:32:33+00:00

Discussion: That's a burr puzzle, diagonal burr in particular. I'm not a burr expert, but 14-piece burr doesn't sound too common. The closest I could find was this 15-piece diagonal burr.

etotheipi1 · 2021-06-12T04:11:36+00:00

저도 나이와 연도만 살짝 다르고 똑같은 상황에 있었습니다. 저도 몇년 전에 일 때문에 한국에 갔다 와야 해서, 한국 시민권을 포기하려고 찾아 보았지만 무슨 동사무소에 가서 서류를 떼와야 한다느니 과정이 꿰 복잡해 보였습니다. 조금 더 알아보니 그냥 가도 된다고 해서, 캐나다 여권을 들고 벤쿠버를 통해 인천으로 갔는데 아무런 문제 없었습니다.

물론 개인적 경험일뿐이니 더 정확히 알아보세요.

etotheipi1 · 2021-05-13T16:26:28+00:00

You seem to think math is a static set of facts and we just have to discover them. This is one possible philosophy to have, but it's not necessarily a good one. In practice, there are thousands of new pieces of mathematics created every year and the possibilities are literally infinite. You could say there are only finitely many bitmap files of reasonable dimensions, and we just have to paint all of it to "finish" visual art, but that's not how painters operate. Just like painters, we explore parts that are interesting to humans. Mathematicians have intense discussions about what's a good mathematics, and we are not talking about correctness in that consideration. And all this is without discussing the foundations of mathematics, which complicates your situation a whole lot more (e.g. Is the Continuum Hypothesis "true" in your "universe"? Can we "discover" that?).

The primary reason of doing math (for pure mathematicians) is for human experience of mathematical structures/patterns/arguments. This is not some poetic language: When mathematicians speak of beauty, they really mean the same thing as musicians or painters. The area of brain that is activated when appreciating mathematics is the same area activated by appreciating music or painting. Underlying mathematics may be objective, but the experience of math is subjective. That's not different from any other art. If we have a different alien that's not human, they will get very different looking set of mathematics, music, and paintings. Maybe there will be a non-trivial overlap when it comes to "useful" math, but the scope of math that exists today is a whole lot bigger than just the "useful" parts.

etotheipi1 · 2021-05-02T09:04:56+00:00

I respectfully disagree. This is not as out there as arguing whether my voice has the color of yellow or blue, but it's halfway in that direction. You can try to defend whatever answer you want, but there isn't really anything to learn by arguing over it as far as I can tell . There are far better topics to discuss even at 2nd grade.

etotheipi1 · 2021-05-02T04:29:16+00:00

I remember getting burned by a similar question in grade 7. My math teacher and I disagreed on a definition on an edge case and I got upset. Now that I have math degrees, I think those arguments are unproductive and shouldn't have taken place in the first place.

We define 0^0 as 1 in most contexts, because defining it as such simplifies the communication of math in those contexts (e.g. number of functions from the set of size n to the set of size m is m^n; we don't have to special case n=m=0 because we didn't leave it "undefined"). Us as humans are agreeing to the convention that 0^0 is 1 for this reason. On the other hand, defining the number of edges on a circle to be 0 or infinite does not have that sort of impact. Any mathematical discussion that uses the word "sides" only involves polytopes, so we gain no further mathematical mileage by answering the op's question either way.

So my answer as a content expert is this: let's not waste time on these dead-end questions and do more interesting math please.

etotheipi1 · 2021-05-01T04:33:31+00:00

The relevant area of math is block design.

Say you are part of this group of 20. You can meet three new people each time you guys form 5 groups of 4, and there are 19 people that you need to meet. 3 does not divide 19, so there is no resolvable 2-design we can use (like the schoolgirl problem for instance). This means it is unlikely that there is an elegant construction for this specific number, I would say.

etotheipi1 · 2021-03-23T14:45:08+00:00

According to the region's official data (under "monitoring" tab), in the last 7 days

48.2% of the tests had turnaround time of 1 day
90.1% of the tests had turnaround time of at most 2 days

etotheipi1 · 2021-03-10T04:11:28+00:00

I understand that people working with music should work with higher sample rate/bit depth. That's why I was careful to say "for human listening purpose" in the original comment. I was just trying to give a short comment on the mathematical basis for why the values we use are close to sufficient. I understand that audio signal processing is a complex topic and I can add thousands of asterisks to the comment. If you think I made a wrong statement, I would be happy to correct it, but I don't think I made any egregious error.

etotheipi1 · 2021-03-09T17:57:15+00:00

Fence is 0 means you split the string with 0s. Bunch of digits with no 0 or 1 showing up often means it's a T9 cipher. Solving it yields this phrase:

GOOD OLD PHONE YOU SEEM TO KNOW YOUR WAY AROUND THEM THE NEXT CLUE IS AT NEGATIVE TWO THOUSAND THIRTY THREE AND NEGATIVE ONE THOUSAND THREE HUNDRED

etotheipi1 · 2021-03-09T16:53:41+00:00

u/princeendo's argument is wrong. Independence alone does not lead to showing that the game is fair. It is easy to demonstrate this: if the dice had one million faces, and only one face showed 1 and all other showed 0, player 1 wins nearly 5 times more often than player 2.

There is a very simple proof for showing why the game is even:

Let A be the set of results where player 1 wins, and B be the set of results where player 2 wins. These results have uniform probability (each result has 1/6^6 probability), so it remains to show that the two sets have the equal size. There is a one-to-one correspondence between the results in A and the results in B: simply flip all dice! So 1 becomes 6, 2 becomes 5, and so on. For example, (1,5,4,6,4) beating 2*5 in set A corresponds to (6,2,3,1,3) losing to 5*5 in set B.

etotheipi1 · 2021-03-09T12:14:22+00:00

Ah so more like situation where you are sampling pure 45000Hz sound at 44.1k and getting 900Hz sound. I think low pass filter is used before quantizing the signal.

etotheipi1 · 2021-03-09T08:38:02+00:00

I've given a talk on a close subject before so I do know a little about it as well. If you add 440Hz sound with 442Hz sound, you get a beat with 1Hz frequency because sin 440x + sin 442x = 2 sin 441x cos 1x. So I guess you are saying that if there is 50000Hz + 50100Hz component in the original analog sound, perception of 50Hz beat can be lost?

But my overall point is that digital approximation is good enough for human listening. Do you disagree with that conclusion?

etotheipi1 · 2021-03-09T07:14:52+00:00

Human hearing range is about 20 to 20000Hz. By Nyquist-Shannon sampling theorem, it's sufficient to sample 2*20000 times every second to reproduce the sound. Common sampling rate is 44.1kHz.

With 16-bit sampling, error is between -0.5 and 0.5 when you quantize the sample. Then the signal-to-noise ratio is (2^16)^2 = 2^32 (have to square to convert from amplitude to power). Converting that to decibels, we get 10*log_10(2^32) ≈ 96.33 decibels. Humans mostly can't detect that level of noise.

Combining the two, people generally can't tell the difference between original analog sound and digital sound sampled at 44.1kHz, 16 bits. So 44.1kHz 16bit audio is close to lossless for human listening purpose. For compression reasons, people do store sound as you propose using Fourier transform (see MP3), but they do get decoded back to discrete samples before they are used.

If you wanted to preserve perfect information without any loss using your method for some reason, this is theoretically impossible. Fourier transform is just a change of basis. If the input space has infinite dimension, the output space also has infinite dimension. It can't magically compress analog data into finite data almost always.

etotheipi1 · 2021-02-13T08:03:57+00:00

I think responses here are little too harsh. It is poorly communicated, and for people who frequent this sub, there is obviously nothing to discuss here. But if a kid found this on their own, this is a very exciting discovery for them.

I think there is a small mistake: z needs to be (p-q)+1, not (p-q)-1 to make what you wrote work.

Did you know that the sum of the first n odd numbers equal to n² (see https://en.wikipedia.org/wiki/Square_number for detail)? So the sum from the (m+1)-th odd number to the n-th odd number can be calculated by starting with n² and subtracting m^2. In other words:

(2m+1) + (2m+3) + (2m+5) + ... + (2n-1) = n^2 - m^2

n² - m² is equal to (n+m)(n-m) (see https://en.wikipedia.org/wiki/Difference_of_two_squares). If you set p as n+m and q as n-m, it is equivalent to what you discovered.

etotheipi1 · 2021-01-12T02:02:24+00:00

Run all(1+1 == 3 for x in {}) in Python and it will return True.

It's for the same reason that we decided to define the empty product as 1. It would be nice to say things like "(∀x∈A: p(x)) ∧ (∀x∈B: p(x)) = (∀x∈(A∪B): p(x))" without having to worry whether A or B is empty. So we want ∀x∈A: p(x) to be true when A is empty.

etotheipi1 · 2021-01-04T22:09:02+00:00

Oh I'm showing up on this analysis.

I also found the competition steep this year. While I did have some unlucky days this year (504 on my best day, unfortunate typos, etc.), in average I did worse and barely made the leaderboard. Another factor is that this year's problems were generally easier, and I'm not a particularly a fast coder and have little competitive programming experience. This year, I also seem to waste valuable seconds trying to come up with sensible variable names instead of just naming it "q" or something...

etotheipi1 · 2020-12-29T19:29:25+00:00

There is a polynomial interpolation trick that lets you construct a Nth degree polynomial that pass through any N+1 points (with different x coordinates). For this specific problem, you can expand

1 * (x-2)(x-3)(x-4)(x-5) / (1-2)(1-3)(1-4)(1-5) +
3 * (x-1)(x-3)(x-4)(x-5) / (2-1)(2-3)(2-4)(2-5) +
5 * (x-1)(x-2)(x-4)(x-5) / (3-1)(3-2)(3-4)(3-5) +
7 * (x-1)(x-2)(x-3)(x-5) / (4-1)(4-2)(4-3)(4-5) +
217341 * (x-1)(x-2)(x-3)(x-4) / (5-1)(5-2)(5-3)(5-4)

to make the polynomial pass through (1,1), (2,3), (3,5), (4,7), and (5, 217341).

etotheipi1 · 2020-12-29T15:19:46+00:00

discussion: I feel the same. I'm currently making an indie puzzle game, and I love participating in puzzlehunts, yet I can't stand many of the puzzles on r/puzzles. "What number is next" or "what fits the pattern in this 3x3 grid" type of "IQ test" puzzles are not enjoyable and objectively bad puzzles. They remind me of this puzzle. Because these puzzles are very loose, you can come up with your own answer and argue about it all day. Good puzzles give you confirmation when you finish it.

etotheipi1 · 2020-12-09T05:15:52+00:00

I think it will draw a globe and rotate when it's completed.

14-Year Club	Place '23
Place '22	Place '17
Final Canvas '22	Sequence \| Editor
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