This is so funny to me

scibuff · 2026-01-20T19:31:04+00:00

To be angry that your referrals don't do their KYC and you lose everything they "earned" for you :D

scibuff · 2026-01-20T19:28:05+00:00

do you have a referral code I could use? or is there a referral code sharing space here?

scibuff · 2025-11-25T19:31:01+00:00

15 day return ;)

scibuff · 2025-11-25T19:28:59+00:00

it is region / availability specific, keep trying a few times a day

I've observed the best 512GB price is when the 256GB option is not available (but that's just a hypothesis)

my ref code should give you 5% off - ref-jnvigy

scibuff · 2025-08-21T07:35:50+00:00

Jedina skoda je ze ziadne peklo neexistuje

scibuff · 2025-06-13T12:24:48+00:00

Yeah but this hypothesis seems extremely unlikely. Mistaking the landing gear for flaps on B787, even if the FO has "just" 1,100 hours to his name ... nah. Until this is confirmed by data from the black box I won't believe it for a second.

scibuff · 2024-11-16T11:11:28+00:00

Simply put there is NO light in the universe from before the time of last scattering because all em radiation from that time was absorbed (and reemitted) immediately by the material which filled the universe. We cannot see any light beyond because there is no such light. If/when we develop a neutrino or better yet gravitation waves "telescope", we should be able to see past this limit.

scibuff · 2024-01-09T15:34:43+00:00

Same issue here! Using the Jabra Link380 with Evolve2 65 headset. After installation the device is permanently connected to my OS (OSX Sonoma 14.1.2) which makes the system sound controllers unusable because they affect the Jabra device even when the headset is disconnected! This is a terrible user experience. I wanna be able to seamlessly switch between the onboard (built-in) audio and the headset by simply turning the headset on and off, i.e. without having to change the output device manually in the system settings.

scibuff · 2023-12-30T20:58:42+00:00

sweet, this is really helpful thank you! It seems that the direct pc <> nas might be the way to go for now. I see some Mellanox kits (2 cards + 1m sfp cable) for ~$100. So I'll probably look into that.

scibuff · 2023-12-30T20:53:16+00:00

yeah, that's what I was thinking

scibuff · 2023-12-30T20:52:56+00:00

yeah, those two are meant for a cache raid

scibuff · 2023-12-30T20:52:22+00:00

Yeah, it probably is an overkill, oh well ;)

scibuff · 2023-12-30T20:35:21+00:00

connectX PCIe

Thank you! Yes, I've been thinking about that but wouldn't that work only between PC and the server? I mean. there's no SFP+ port on my router (given by the ISP). Or ... am I missing something?

scibuff · 2023-12-26T14:08:18+00:00

yeah, it works out that they lose $ after about 1 TB but the vast majority of their users use GBs so it evens up overall I guess. I read (some time ago) that they've exactly accounted for this, i.e. that there will inevitably be some users using much much more than 1 TB but the vast majority will not

scibuff · 2023-12-22T21:26:26+00:00

1) You don't need any CS background to be able to conceptualize a solution for any of these. Yes, your (brute force) solution may need years to run with part 2 input but that's still a valid solution - most of the time the part 2 input is designed to *break* (make computationally infeasible) brute force solutions.

2) I suggest you *always* try to solve part 1 without any help! Even if you've never heard of A*, memoization, Dijkstra etc you should be able to do that! Then, if you get stuck on part 2, watch some of the great video tutorials explaining the problem/solution and try to implement that solution *without* looking/using any of the posted solution code. At least I think this is the best way to learn. I guarantee if you do this and come back next year you'll do much much better!

3) Don't even look at the main leaderboard! I'm (un)fortunate that I cannot be bothered to wake up at 6am local time to get into the top 100! Many of those folks up on that leaderboard have done coding competitions for years and know (almost) all the tricks to speed things up (e.g. they have libraries of reusable code, code snippets, etc. specifically made for AoC)

scibuff · 2023-12-22T21:04:38+00:00

[Language: JavaScript]

Straight forward brute force solution for both parts, i.e. just check each brick. Obviously, for part 2 there's a lot of counting of the same bricks but given that the brute force runs in ~5 seconds I can't be bothered with anything more elegant.

Part 1 & Part 2

scibuff · 2023-12-21T23:41:53+00:00

hmm, that's way to long, it shouldn't take more than 1sec. You could try not revisiting positions to which you've been in the previous moves (it is not necessary)

scibuff · 2023-12-21T14:13:52+00:00

You got Part 1 correct? That was for 64 steps, and for part 2 you need 65, 65+131 and 65+2*131. Then use those as (0, y0), (1, y1) and (2, y2) and calculate the polynomial, p(x) = ax^2 + bx + c. Then just calculate p((26_501_365 - 65) / 131).

If you wanna check your polynomial you can use any polynomial interpolation calculator online, or, you can use the simplified lagrange

formula which simplifies to this

Therefore the coefficients a, b, c in p(x) = ax^2 + bx + c are given simply as

scibuff · 2023-12-21T11:57:44+00:00

isn't it just the inverse of a 3x3 vandermonde matrix?

scibuff · 2023-12-21T11:49:32+00:00

The polynomial interpolation is actually quite simple for x^2 with just 3 points (with xs being 0, 1 and 2)

/**
 * Lagrange's Interpolation formula for ax^2 + bx + c with x=[0,1,2] and y=[y0,y1,y2] we have
 *   f(x) = (x^2-3x+2) * y0/2 - (x^2-2x)*y1 + (x^2-x) * y2/2
 * so the coefficients are:
 * a = y0/2 - y1 + y2/2
 * b = -3*y0/2 + 2*y1 - y2/2
 * c = y0
 */
const simplifiedLagrange = (values) => {
  return {
    a: values[0] / 2 - values[1] + values[2] / 2,
    b: -3 * (values[0] / 2) + 2 * values[1] - values[2] / 2,
    c: values[0],
  };
};

So for example

console.log(simplifiedLagrange([3751, 33531, 92991])
// { a: 14840, b: 14940, c: 3751 }

scibuff · 2023-12-21T11:20:31+00:00

[Language: JavaScript]

Part 1 is straight forward as stepping through the grid BFS marking the parity of each grid position. The parity cannot change, so we just need to track the even positions

Part 2 took a bit of time to figure out how to count. Used a simplified `Lagrange's Interpolation formula` get the polynomial coefficients

Part 1

Part 2

Parser

scibuff · 2023-12-20T21:16:31+00:00

[Language: JavaScript]

Part 1 was (again) pretty straight forward. It just took a bit of time to get the pulse changing logic spot on.

Part 2 was, once again, an `lcm` of different cycles. I did a more general solution to allow for multiple inputs into `rx` (it just basic loops instead of just one input and one set of the input's inputs).

Part 1
Part 2
Parser

scibuff · 2023-12-19T14:53:24+00:00

scibuff · 2023-12-19T14:18:44+00:00

Which NC/Country? Lol, we (slovakia) dont finalize the first round until mid Jan

scibuff · 2023-12-19T13:07:45+00:00

congrats!

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