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

Advent Of Code

770 readers
75 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 1 year ago
MODERATORS
 

Day 12: Hot Springs

Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • Code block support is not fully rolled out yet but likely will be in the middle of the event. Try to share solutions as both code blocks and using something such as https://topaz.github.io/paste/ , pastebin, or github (code blocks to future proof it for when 0.19 comes out and since code blocks currently function in some apps and some instances as well if they are running a 0.19 beta)

FAQ


๐Ÿ”’ Thread is locked until there's at least 100 2 star entries on the global leaderboard

๐Ÿ”“ Unlocked after 25 mins

you are viewing a single comment's thread
view the rest of the comments
[โ€“] mykoza@lemmy.world 2 points 11 months ago (1 children)

I'm struggling to fully understand your solution. Could you tell me, why do you return 1 when at the end of a and b ? And why do you start from size + 1?

[โ€“] cvttsd2si@programming.dev 1 points 11 months ago (1 children)

T counts the number of ways to place the blocks with lengths specified in b in the remaining a.size - ai slots. If there are no more slots left, there are two cases: Either there are also no more blocks left, then everything is fine, and the current situation is 1 way to place the blocks in the slots. Otherwise, there are still blocks left, and no more space to place them in. This means the current sitution is incorrect, so we contribute 0 ways to place the blocks. This is what the if bi >= b.size then 1L else 0L{.scala} does.

The start at size + 1 is necessary, as we need to compute every table entry before it may get looked up. When placing the last block, we may check the entry (ai + b(bi) + 1, bi + 1), where ai + b(bi) may already equal a.size (in the case where the block ends exactly at the end of a). The + 1 in the entry is necessary, as we need to skip a slot after every block: If we looked at (ai + b(bi), bi + 1), we could start at a.size, but then, for e.g. b = [2, 3], we would consider ...#####. a valid placement.

Let me know if there are still things unclear :)

[โ€“] mykoza@lemmy.world 1 points 11 months ago (1 children)

Thanks for the detailed explanation. It helped a lot, especially what the tbl actually holds.

I've read your code again and I get how it works, but it still feels kinda strange that we are considering values outside of range of a and b, and that we are marking them as correct. Like in first row of the example ???.### 1,1,3, there is no spring at 8 and no group at 3 but we are marking (8,3) and (7,3) as correct. In my mind, first position that should be marked as correct is 4,2, because that's where group of 3 can fit.

[โ€“] cvttsd2si@programming.dev 1 points 11 months ago (1 children)

If you make the recurrent case a little more complicated, you can sidestep the weird base cases, but I like reducing the endpoints down to things like this that are easily implementable, even if they sound a little weird at first.

[โ€“] mykoza@lemmy.world 2 points 11 months ago

You are probably right. Just my rumblings. Thanks for the help.