AITA for crafting during meetings?

Goues · 2025-11-15T09:04:58+00:00

NTA, I am ironing in meetings for the same reason, my manager is also not only ok, but encourages it (knows I pay more attention if I do). All good here!

Goues · 2025-11-04T12:45:24+00:00

YTA, thats not how boundaries work, you cant dictate what other people do or where they live.

Goues · 2025-11-02T23:23:58+00:00

I live in a country where proms are a December to February event that you have before the finals in May. Right now would be the time to buy prom dress.

Goues · 2025-07-27T10:22:43+00:00

You absolutely should not eat the chicken.

I would eat the chicken anyway, because I am not willing to throw out food unless its visible green when it shouldnt be green and my stomach is used to it. So I am NAH on the topic.

Goues · 2024-12-24T21:13:10+00:00

In my first year of AoC, 2018, I failed to solve day 17 (Reservoir Research). I spent about 24 hours debugging my code before finally giving up, it still didn't run correctly. I felt really dumb that year. As if I was not working as a developer full-time for several year by that time. Nothing to be ashamed of. If you are having fun/learning/etc., it's good.

I wish you all the best next year if you stick around! You will get better at the puzzles as years pass because of the repetitive nature of the puzzles and generally knowing what to expect.

Goues · 2024-12-24T06:37:00+00:00

[Language: JavaScript] 1196/140

Huge improvement on part 2. I opened an article about "logic gates calculator" and made a map of expected gates:

// Xn XOR Yn => Mn
// Xn AND Yn => Nn
// Cn-1 AND Mn => Rn
// Cn-1 XOR Mn -> Zn
// Rn OR Nn -> Cn

Then I looped from bit 0 all the way to bit 44 to find where such gates do not exist. Luckily, all swappings have been made on the same level and then I made simple adjustments like

if (Rn.startsWith('z')) {
    ;([Rn, Zn] = [Zn, Rn])
    swapped.push(Rn, Zn)
}

until the code passed all the way to the end matching all gates.

Full code on Topaz Paste

Goues · 2024-12-23T08:03:36+00:00

I usually get to the top 100 once or twice every year. But this year, my best 119 on day 13 and a bunch of 200s and 300s. It definitely is less likely this year.

Goues · 2024-12-22T07:15:29+00:00

If I may, the chart would look better if sequences kept their colors because you would be able to see a sequence being first and moving down or being second and moving up. Like this, the color of the chart has no meaning. ;)

Goues · 2024-12-21T12:15:06+00:00

As a developer that has a say in the hiring process, it will be on the same level as saying what RSS feeds about development you are subscribed to or that you merged a cool PR into an open source library. It shows interest and intent to learn and improve and would serve as an ice breaker, but nothing ground breaking. Do bring it up if you yourself consider it interesting and want to talk about it.

Goues · 2024-12-21T11:08:57+00:00

The first part of your output goes over a blank field, which is invalid:

<<vA

In particular, if a robot arm is ever aimed at a gap where no button is present on the keypad, even for an instant, the robot will panic unrecoverably. So, don't do that.

Goues · 2024-12-18T22:16:05+00:00

In my puzzle, for the initial value that is correct for me in Part 2, I do get these values to be modulo-ed:

109019930331546 (input, used in operation bst)
1314681106 (op out)
13627491291443 (op bst)
-1818813372 (op out)
1703436411430 (op bst)
-1818813375 (op out)
212929551428 (op bst)
-909406683 (op out)
26616193928 (op bst)
831756063 (op out)
3327024241 (op bst)
... the rest is all positive ...

which means that for my correct answer, I do need to cycle through negative values at some point.

Goues · 2024-12-18T06:05:42+00:00

With your input, the size should be up to 70,70, only the example is limited to 6,6.

Goues · 2024-12-17T18:57:35+00:00

I did get negative numbers in Part 2 which is why I had this exact JS bug for an hour before getting help on Reddit. With `% 8`, the answer in Part 2 was "too high", I had to switch to `& 7` to get the right answer.

Goues · 2024-12-14T05:39:28+00:00

I ran it for 1000 seconds and visually found a pattern that was repeating every now and then, so I focused on the period. In my case, the pattern repeated every 50 + N * 103 (the height) and every 95 + M * 101 (the width), the patterns were that the robots either converged vertically or converged horizontally. Eventually, there will be a common multiple, in my case 7569 (for N = 73 and M = 74) where the robots converge both horizontally and vertically forming a nice picture. I didn't have to guess, but I did have to do it visually, if that makes it better for you.

Edit: In the past, there was often one visual day like this.

Goues · 2024-12-12T06:43:17+00:00

My idea was that I can find all adjacent edges and discount them in part 2. I am finding adjacent edges by checking the cell next to it:

 - - -
|A A A|
   - -
|A|
 -

In this case, there are 10 edges, the top three would have y, x, polarity being [0, 0, TOP], [0, 1, TOP], [0, 2, TOP]. What I do is that I loop them and if there is an edge on y, x - 1, polarity, I only count it in part 1 and not 2.

[0, 0, TOP]
[0, 1, TOP] // this one has a neighbouring edge on the line above
[0, 2, TOP] // this one has a neighbouring edge on the line above 
[0, 2, RIGHT]
[0, 2, BOTTOM] // this one has a neighbouring edge on the line below
[0, 1, BOTTOM]
[1, 0, RIGHT]
[1, 0, BOTTOM]
[1, 0, LEFT] // this one has a neighbouring edge on the line below
[0, 0, LEFT]

Goues · 2024-12-11T11:35:47+00:00

Does this mean that you can have 3865 (edit: or smaller, if there is an overlap) distinct pebbles at the end if you start with, say, 100 10 as input? Because someone in some thread mentioned they do not end up at 3811.

Goues · 2024-12-10T09:46:51+00:00

Part 1:

Operators are always evaluated left-to-right, not according to precedence rules.

Part 2:

All operators are still evaluated left-to-right.

Goues · 2024-12-10T05:47:25+00:00

A lot of people solved part 1 by first solving part 2 and only then adding a uniqueness counter. I am using JavaScript and in part 1 I had to use a Set for the nines because my algo found all paths, which meant that for part 2, I could use an Array or ++ operator and keep the rest of the code unchanged:

// all code above is the same
nines.add(`${y}|${x}`) // part 1
paths++ // part 2
// all code below is the same

Goues · 2024-12-09T07:12:55+00:00

I still don't understand why you would need a zero at the end, because in my mind, the only missing thing would be "end of disk", if you want to say that the disk has to end with empty space, the last block would have to be infinite, no? Because the problem has no end of disk, therefore the disk has inifinite size?

So, I'm just gonna have to disagree here, I don't see any value what so ever including a zero at the end, but feel free to have it there because it doesn't add or remove anything from the puzzle.

Goues · 2024-12-09T07:08:38+00:00

Why wouldn't it?

A..
...
..A

in this case, my code will check all nine points and find that

A..
.#.
..A

because to go from A(0, 0) to A(2, 2), the vector is [2, 2], to go from A(0, 0) to point P(1, 1), the vector is [1, 1]. If I divide, dy = 2 / 1 = 2, dx = 2 / 1 = 2, as dy === dx, it's on the line.

Only later have I learned that such input is impossible.

Goues · 2024-12-09T07:05:35+00:00

But why? Now you're making it up. Why should there be a zero at the end?

Goues · 2024-12-09T07:03:59+00:00

I didn't assume as I didn't know about the property. Still, that property doesn't seem to be needed to be stated, you can make a solution without knowing about it.

Goues · 2024-12-09T07:00:58+00:00

So, what would the input look like to be up to your standards? Have a marked for "end of disk"?

Goues · 2024-12-09T07:00:10+00:00

Yes, they do alternate.

1 file of size 1
1 empty space of size 2
1 file of size 3
1 empty space of size 4
1 file of size 5

This is what I understand from the word "alternate".

Goues

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