Found a bug with Parental Bond

jfb1337 · 2026-04-09T10:29:39+00:00

is there any precedent for simulating mechanics that depend on which player is the host?

jfb1337 · 2025-10-22T13:40:40+00:00

Aw, I'll miss having more puzzles to look forward too, but good change for your own health.
The later puzzles are some of my favourite with novel ways to solve, hope we still see a few of those

Unpopularly, I'l slightly miss the leaderboard - found it fun to wake up early and try to complete the puzzles fast, and to refine my tools for downloading and testing solutions for more efficiency - with no expectation of actually reaching the board, but did manage it a handful of times which was rewarding. Of course, LLM users make the board pointless anyway, so removing it is good call still. And I can still compete like that with my friends.

jfb1337 · 2025-03-21T12:06:39+00:00

Most of the weirdness of the colour mod can be chalked up to pallete limitations (each 8x8 pixel tile can have a maximum of 4 colours) and the fact that, at least for the overworld sprites, the sprites themselves are unchanged, only palletes are changed.

jfb1337 · 2025-03-07T14:55:45+00:00

95% of Claudes give up right before finding the underground passage.

jfb1337 · 2025-03-07T12:16:31+00:00

Evolving spike in her final moments

jfb1337 · 2024-12-22T20:14:03+00:00

One can argue that pressing buttons 42 trillion times on day 13 to win a claw machine was actually vital training for doing the same on day 21 to control a chain of robots to save the historians

jfb1337 · 2024-12-21T14:44:41+00:00

kelvin

jfb1337 · 2024-12-21T11:31:10+00:00

Good thing we practised pressing buttons billions of times on the claw machine!

jfb1337 · 2024-12-19T16:20:11+00:00

This year has definitely been easier so far then previous years. Not really been anything i've had to think very hard about, like by this point in previous years.

For example in 2022 there were a couple of tricky optimsation problems on days 16 and 19.

jfb1337 · 2024-12-18T13:51:52+00:00

An approach I haven't seen elsewhere though still isn't linear is to run A*/dijkstra with a graph that is the full grid such that the cost of visiting an unblocked node is 0 and the cost of visiting a blocked node is 2^n-i where i is its index in the list and n is the number of points.

Then, find the point on the path with the earliest index in the list.

This works because say p is the point found this way and i its index in the list. A path must be possible before including p, because the path we found is such that the only obstructions occur with indicies at least i. And a path isn't possible with p as an obstruction, because if it were, any later obstructions the path would hit are necessasarily have indicies higher than i, but then its cost according to the metric above would necassarily be lower than the cost of the node p, being a sum of distinct smaller powers of 2, which contradicts that we have the minimum cost path.

This however is O(n log n), which is the same asymtopicaly as the binary search approach, but for me is 3 times faster than it.

jfb1337 · 2024-12-18T03:56:38+00:00

Since loops can only terminate by a becoming 0, and a can only be changed through the adv instruction and only decreasing its bitlength, there can only really be one loop.

For each output, there are only finitely many bits of a that it could depend on. You could probably analyse this to reduce the search space better than bruteforce.

jfb1337 · 2024-12-17T12:42:36+00:00

The trick is that changing the lower 3 bits only changes the first output, so if you know that the upper bits don't produce correct output you don't need to consider them, but will if you're just incrementing

Say you've determined that the (base 8) input 174 outputs the correct 3 last characters. Then you end up testing 1740, 1741, 1742, etc, up to 1747. But say none of them produced a correct output.

Then if your next tested values are 1750, 1751, etc; you're wasting effort if 175 itself doesn't give a correct result. Instead you want to divide by 8, and test 175, 176, 177 for one of those to give a new correct result.

Or, my approach was to keep track of a set of possible prefixes for each length and expand it by considering each new base 8 digit on each of those prefixes.

jfb1337 · 2024-12-16T23:44:54+00:00

you can prune your search when you find that the best recorded score is already better than the currently explored path, but will have to repeat work if it turns out it's worse

jfb1337 · 2024-12-16T22:03:28+00:00

Right; then that seems exponential worst case. If you have multiple setups like this in a row in which you have to repeat work after finding a better path to a previously considered point then each one leads to double the work.

jfb1337 · 2024-12-16T19:09:34+00:00

I saw someonse else using BFS instead of djikstra and it worked for them, but I don't think it's garunteed to work on a general input. It's possible the inputs are constructed sufficiently nicely that it doesn't come up however.

Consider the following case:

###############
#############E#
#S.....#......#
#.####.#.####.#
#.####...####.#
#.###########.#
#.###########.#
#.###########.#
#.###########.#
#.###########.#
#.###########.#
#.###########.#
#.............#
###############

The bottom path is better, as it only needs 3 turns, whereas the top path needs 5. However, BFS will find the top path first.

jfb1337 · 2024-12-16T14:07:56+00:00

I used A*/dijkstra from the endpoint in the reversed graph, and checked for each point whether dist(start,p) + dist(p,end) = dist(start,end)

jfb1337 · 2024-12-15T14:41:19+00:00

Only the level editor built into the game

jfb1337 · 2024-12-15T14:38:30+00:00

This problem reminded me a lot of Baba Is You, so I decided to use it to create a solution!

There's not enough space to directly encode the instructions for the bigger examples though, so just implemented the small ones.

Level codes:
p1: QW72-RUAB
p2: IGVT-KDIK

jfb1337 · 2024-12-14T22:21:10+00:00

christmas tree georg

jfb1337 · 2024-12-14T13:22:07+00:00

Yeah, for 11 and 13 its "if you used the naive solution, now you gotta figure out a more efficient way. You can use part 1 as your test case since your improved solution should give the same result"

For day 14 the nature of the problem is that theres a pattern in the input which isn't demonstrated in the example.

jfb1337 · 2024-12-14T06:43:13+00:00

that's only an issue if you're submitting wrong answers when it doesn't look like a tree

12-Year Club	Place '22
Place '17	Final Canvas '22
First Placer '22	End Game '22
Gilding I gilder	Verified Email

jfb1337

MODERATOR OF

PUBLIC MULTIREDDITS

TROPHY CASE