this post was submitted on 12 Dec 2023
6 points (100.0% liked)

Advent Of Code

158 readers
1 users here now

An unofficial home for the advent of code community on programming.dev!

Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

AoC 2023

Solution Threads

M T W T F S S
1 2 3
4 5 6 7 8 9 10
11 12 13 14 15 16 17
18 19 20 21 22 23 24
25

Rules/Guidelines

Relevant Communities

Relevant Links

Credits

Icon base by Lorc under CC BY 3.0 with modifications to add a gradient

console.log('Hello World')

founded 11 months ago
MODERATORS
 

So I managed to get part 1 of the day, but it took 2 seconds to run on the real input, which is a bad sign.

I can't see any kind of optimisation that means I can skip checks and know how many combinations are in those skipped checks (aside from 0.) I can bail out of branches of combinations if the info so far won't fit, but that still leads me to visiting every valid combination which in one of the examples is 500k. (And probably way more in the input, since if I can't complete the example near instantly the input is not happening.)

Right now I take the string, then replace the first instance of a ? with the two possible options. Check it matches the check digits so far then use recursion on those two strings.

I can try to optimise the matching, but I don't think that solves the real problem of visiting every combination.

I don't think (or hope) it's just bad code but this is my code so far (python.)

edit:

spoilera cache was the solution!

you are viewing a single comment's thread
view the rest of the comments
[–] [email protected] 3 points 6 months ago (1 children)

I can bail out of branches of combinations if the info so far won’t fit, but that still leads me to visiting every valid combination which in one of the examples is 500k.

By "every valid combination" do you mean every substitution of '?' with a '#' or '.'? If yes, then you're wrong, you can bail out of branches that don't fit early, and cut a lot of them this way.

Consider the following example:

???????????? [1, 2, 2]

When you substitute the first two question marks with ##, the answer already doesn't match the input string, so you can throw away 1M of the combinations that don't fit.

Also, while you're at it, avoid generic type annotations (e.g. list), try to always specify the generic argument (e.g. list[str]) :)

[–] [email protected] 2 points 6 months ago* (last edited 6 months ago)

When you substitute the first two question marks with ##, the answer already doesn’t match the input string, so you can throw away 1M of the combinations that don’t fit.

You know I figured I was already doing that, but printing your example shows I was not. I also added some logic to the other end since 1,2,2 needs a space of 7 and if the check is all dots and I only have 6 chars left, I know it can't fit.

still taking a long time for the real data so it must be something inefficient in my code then, rather than the method.

Also, while you’re at it, avoid generic type annotations

Good point. Recently figured that one out, still not automatic as you can see.

Thanks for the pointers.