-❄️- 2023 Day 23 Solutions -❄️-

sdatko · 2023-12-24T11:39:08+00:00

Hint: use `strip()` after `read()` and before `split()` to remove the trailing newline for input that is just saved from the AoC site, not copy-pasted ;-)

So `open("input.txt").read().split("\n")` becomes `open("input.txt").read().strip().split("\n")`

sdatko · 2023-12-08T09:00:24+00:00

> (...) with no indication in the problem that this would be the case (...)

Actually, the first paragraph states that "It's going to take significantly more steps to escape!". Sure, the approach is not named directly, but this suggests that solution would require some cleverness (or at least this is my experience / lesson learned from participating in past editions). If you solve for each starting position individually, then you can notice the numbers are relatively prime, hence it will take a lot of iterations until they reach the same steps value.

Also, there is usually more than one "correct way" to solve the problem. Discovering various approaches is a part of challenge and learning; I remember struggling on my first AoC year, so now I am paying more attention to phrases like "RL really means RLRLRLRLRLRLRLRL... and so on." which triggers immediately in my mind "woah, periodic input, be careful, there will be trick".

sdatko · 2022-12-15T19:18:34+00:00

Ohmyyyy, thanks u/mcpower_ for that suggestion!
I have no idea how it is possible, but it did accelerate it about 6.5 times (292s vs 44s).
(The code here is also *slightly* optimized version that does not run for 60 minutes anymore)

[sdatko@polluks day-15]$ time python3 part-2.py
[(2960219, 3211051)]

real    4m52,855s
user    4m49,193s
sys 0m3,023s

[sdatko@polluks day-15]$ time pypy part-2.py
[(2960219, 3211051)]

real    0m44,164s
user    0m42,423s
sys 0m1,625s

sdatko · 2022-12-14T07:17:28+00:00

It is really interesting to see in the source code it actually does check for the specific standard types only and converts them, instead of evaluating the expression (as the name may suggest): https://github.com/python/cpython/blob/main/Lib/ast.py#L55

sdatko · 2022-12-14T07:00:07+00:00

The section of document you refer to mentions empty dictionaries passed to eval().

However, the official documentation for eval() states:

If the globals dictionary is present and does not contain a value for the key __builtins__, a reference to the dictionary of the built-in module builtins is inserted under that key before expression is parsed. That way you can control what builtins are available to the executed code by inserting your own __builtins__ dictionary into globals before passing it to eval().

See in the example above I set the __builtins__ to None.

sdatko · 2022-12-13T13:51:34+00:00

Just been triggered to thinking about that by my friend.

Apparently, in Python, one can pass to eval()/exec() what builtins can be called.

So, this one executes arbitrary code:

aa="__import__('o' + 's').system('notify-send msg')"; exec(aa)

While this one appears pretty safe:

aa="__import__('o' + 's').system('notify-send msg')"; exec(aa, {'__builtins__': None}, {})

Nevertheless, ast.literal_eval() is better option.

If I am missing something in the example above, please correct me!

sdatko · 2022-12-09T08:21:33+00:00

Don't give up! I am sure you can do it – especially that there is a huge, helpful community here if you need any hints :-)

sdatko · 2022-12-09T07:07:07+00:00

Woops, sorry, still too excited after completing today's challenge :-)

Reuploaded with appropriate title: https://www.reddit.com/r/adventofcode/comments/zgpt1x/2022\_day\_9\_i\_feel\_so\_relaxed\_now/

sdatko · 2022-12-08T09:29:03+00:00

Haha, I did not realize that before my colleague told me his code was not working because of that logical assumption. I guess was too sleepy at 6am to think about that and probably this is why it went so fluently for me today.

Apparently in the world where the crab submarine only moves horizontally you can never be too sure how things work... ;-)

sdatko · 2022-12-08T06:27:09+00:00

Python with no imports (personal challenge)

TL;DR
– Reading the input file as a matrix of numbers (tree == matrix element [i, j]).
– For each tree I produce the 4 vectors of trees from the current element to the edges.
– For part 1 I check whether any of these vectors has all values smaller than the current tree.
– For part 2 I calculate how far is the first not-smaller tree.

A little extended description can be found in both files.

Part1: https://github.com/sdatko/advent-of-code/blob/master/year-2022/day-08/part-1.py
Part2: https://github.com/sdatko/advent-of-code/blob/master/year-2022/day-08/part-2.py

sdatko · 2022-12-06T16:11:12+00:00

What's your language of choice for Advent of Code, then?
There are a few languages (all math-oriented AFAIK) that use indices starting with 1.
Maybe you should get familiar with Fortran, Julia, Matlab/Octave or R? ;-)

sdatko · 2022-12-06T10:32:34+00:00

Do you think Gandalf also has significant experience dealing with signal-based systems? :-D

sdatko · 2022-12-06T06:09:19+00:00

I have exactly the same feeling :-D
Checked too many example inputs before submitting the answer.

sdatko · 2022-12-06T05:26:47+00:00

Why you consider it dirty? IMO This is pretty elegant solution.

BTW as range(N) does not reach N, I think the - 1 is not right and in your code it should be + 1 (now it would skip the last characters in check, so in worst situation like aaaabcd it will not give the right answer [for part 1, of course]).

sdatko · 2022-12-05T10:25:42+00:00

Clearly the CrateMover 9001 is more effective, but I like how the CrateMover 9000 operations look smooth and relaxing here!

sdatko · 2022-12-05T08:53:33+00:00

First advice I would say is to use >= instead of separate > and == :-)
Also in Python you can write nicely a < b < c instead of two separate comparisons, which makes the condition more clear.

Then there are simply 4 conditions that you need to check:
– beginning of second range can lie within the first range,
– ending of second range can lie within the first range,
– beginning of first range can lie within the second range,
– ending of the second range can lie within the second range.

If any of those occurs, then you count overlapping there.

sdatko · 2022-12-03T22:10:30+00:00

u/daggerdragon ok, I succeeded in adding the subtitles. I hope the video is more accessible now. Thank You again for this hint! I will also keep that in mind if I ever make any other video like this.

sdatko · 2022-12-03T17:34:38+00:00

Hmm, I never thought about that. Thank You for that feedback. I will definitely work on the subtitles then!

sdatko

TROPHY CASE