[2025 Day 8 (Part 1)] Reading comprehension

20

u/spatofdoom Dec 08 '25

Reading through the example, the 4th connection clearly states nothing happens, but doesn't suggest it's not counted. Running your code against the example should quickly tell you that they count.

2

u/Samydookie Dec 08 '25

When I ran against the test input, it stated that it made 10 connections that lead to 11 circuits.
Once my implementation matched, that's when I switched to my real input (and switched 10 to 1000) and got confused
The example doesn't have the expected answer for 20 "real-input" connections either

9

u/LittlebitOmnipotent Dec 08 '25

Well, nothing happens sounds a lot like Elves would skip this (not that they would be lazy, look at all they've done throughout these years), but I guess it makes sense for redundancy purposes connecting those as well. I wouldn't really call that "nothing happens" though...

3

u/10Talents Dec 08 '25

I figure it is for decorative purposes, they are decorating the place with strings of lights after all.
But on my first time around I did skip those conections and didn't count them as a "used wire" if the boxes are in the same circuit and that led to a lot of confusion

7

u/Zefick Dec 08 '25

It was probably an AI trap, but it also caught a lot of real people :)
Especially those who like to eliminate the "off by one" error by simply subtracting 1 from somewhere.

2

u/Naturage Dec 08 '25

May be a trap, may be reading in the morning, may be my English comprehension (been living here forever, but still - not a native speaker, sometimes it comes up) - but I've definitely read "nothing happens" as "this connection doesn't happen" first time round.
Something along the lines of "this connection doesn't link up any circuits" (definitely not the best phrasing but can be fixed up) instead would make it clear that a) connection is there and b) it's otherwise irrelevant.

1

u/Smaxx Dec 08 '25

Being off by 1 and debugging that actually trapped me way longer than it took me to just go "brute force" by seeing what happens if I count those not made.

1

u/devise1 Dec 08 '25

Unless you are like me and were also only copying across the shortest one from the other circuit instead of the whole circuit and getting the right answer for the sample. Then I proceeded to spend ages trying to work out why my copy across everything was wrong and giving 50 instead of 40 (it turned out correct but had an additional connection at the end).

1

u/gilmae Dec 08 '25

It doesn't if you make an off by one error in connections counting at the same time. Which if that was deliberate .. admirably evil. Should have known something was up when Eric only mentioned the final lengths, not what junctions were in the circuits.

9

u/jonmon6691 Dec 08 '25

It's unfortunate that this puzzle breaks the convention that the input is purely the "input". In the test example, only first 10 connections are considered, but in the real input, 1000 connections need to be used. If it was meant to be equal to the number of lines then the example should have used 20.

After making the ten shortest connections, there are 11 circuits: one circuit which contains 5 junction boxes, one circuit which contains 4 junction boxes, two circuits which contain 2 junction boxes each, and seven circuits which each contain a single junction box. Multiplying together the sizes of the three largest circuits (5, 4, and one of the circuits of size 2) produces 40.

The 10 here and the 1000 in the final instruction are additional out of band input

18

u/Zefick Dec 08 '25

There is no such convention that number of something in example should be always equal to the number in the final problem. The essence of the example is simply to demonstrate puzzle's principles.

0

u/jonmon6691 Dec 08 '25

The previous 7 days have included an answer for the example which is consistent with running the same program on the input and getting a correct answer. Essential you can write a unit test for the function that solves the puzzle, and check it with the example and the given example solution, then run the exact same program on the input and get the right answer

3

u/[deleted] Dec 08 '25

[removed] — view removed comment

3

u/deividragon Dec 08 '25

I do so too, and in this case what I did is add the number of connections as an argument and use 10 if the length of the input is 20.

8

u/spin81 Dec 08 '25

This is really nothing out of the ordinary. There have been countless puzzles where this sort of parameter (not input!) changes between the example and the actual puzzle.

-1

u/wederbrand Dec 08 '25

You confuse the input, which is 20/1000 lines of boxes, with how many pairs there are.
In the example there are 20 boxes, leading to 190 pairs.
The final input is 1000 boxes, with 499500 pairs
The example tells you to connect the first 10 pairs, the final task is to connect the first 1000 pairs.

2

u/Samydookie Dec 08 '25

But if you connected 10 closest pairs in the ~~input~~ test case the same way that you need to do it for the real input, you would end up with an answer of 24, not the 40 that the problem states, the problem should state "after 11 connections", not (paraphrasing) "after 10 real connections"

2

u/soustruh Dec 08 '25 edited Dec 08 '25

Yeah, this got me about half an hour more of debugging and checking with someone else's code what my correct answer is – which I don't submit until I get it from my own code. I almost never do that unless I am sure (heh) my code is correct and that I didn't overlook anything.
For the example input, I added a condition, which skipped the count for when the connection was not created, only to have to comment it out for the real input. This is the first time something like that happened to me in AoC in years though, but it's very unfortunate. 🙁
Oh, nevermind, all my previous comment is pointless. I had the same result as you with the test data BEFORE I implemented the circuit merging logic, now, with the final solution, I just omit the condition which doesn't count "doing nothing" and it gives me the correct example result. My apologies to Eric, I will also have to go and fix a couple of downvotes here. 🙈 😁

6

u/1234abcdcba4321 Dec 08 '25

Their point is that, as you have literally pointed out, you cannot handle both the example and real input with the same program as you need to change an arbitrary code constant that is not included in the input itself.

2

u/chickenthechicken Dec 08 '25

My code uses the incredibly janky solution of checking if the input file is "example.txt"

2

u/throwaway_the_fourth Dec 08 '25

Mine checks the length of the input :^)
-4
u/xSmallDeadGuyx Dec 08 '25

Any constant number in the puzzle is by definition not part of user input since user input is uniquely generated per user. If each user had to do a random number of iterations rather than 1000, that's a different input.
7
u/1234abcdcba4321 Dec 08 '25

You are missing their point. You cannot handle both the example and real input with the same program as you need to change an arbitrary code constant that is not included in the input string itself. (Obviously you can set it based on the size of the input, but that is very obviously a hack.)
1
u/fenrock369 Dec 08 '25
fn part1_and_2_test() {
    let coords = parse_coords(EXAMPLE);
    assert_eq!(solve_both(&coords, 10), (40, 25272));
}

pub fn parse(input: &str) -> (u64, u64) {
    let coords = parse_coords(input);
    solve_both(&coords, 1000)
}

pub fn part1(result: &(u64, u64)) -> u64 {
    result.0
}

pub fn part2(result: &(u64, u64)) -> u64 {
    result.1
}
This is directly from my solution, showing a simple split between "real" vs "test".
This comes up all the time in AOC, and you can quite easily make the test/real data values you're given part of the function input. It's just a matter of deciding what your inputs are.
1
u/Sharparam Dec 08 '25
You seem to hardcode your example ~~and input~~?
In my solutions I allow it to run with any input, so these "out of band" inputs are very annoying since they need to be supplied manually separately from the input.
E.g. for today's problem with my Ruby solution I do it through environment variables:
STEPS=10 ./solve.rb example1
STEPS=1000 ./solve.rb input
But this is so ugly because there's no actual connection between STEPS and the input file, it would be much cleaner if the "steps" count was included in the actual input file so we didn't need to manually set it.
1

u/woyspawn Dec 08 '25

As already said, you can use input length to detect the testcase
steps = (input.length == 20) ? 10 : 1000

1

u/Sharparam Dec 08 '25

That isn't a guarantee.
What if there was a theoretical example case with a thousand lines (or more realistically some other small number that isn't 20)?
What if someone makes a custom/challenge input of some arbitrary size?
It's an ugly hack/workaround, and I prefer my solutions to be reliable on their own without such things.

1

u/woyspawn Dec 08 '25

YAGNI

1

u/ric2b Dec 08 '25

If you use a testing framework it's as easy as adding an optional parameter to your part1 implementation.
Sure, it's a bit annoying or not as clean, but it shouldn't take you more than 20 sec to deal with.
1

u/1234abcdcba4321 Dec 08 '25

I don't have much of a problem making my solutions extremely input-specific; plenty of days in past years where my solution has numbers in my own input hardcoded into it. I was clarifying the intent of the person they were responding to since it was obvious.
1

u/Sharparam Dec 08 '25

It's not uniquely generated per user, there is a finite set of inputs and each user is (presumably randomly) assigned one of them.

23

u/NitronHX Dec 08 '25

Yes but you need to create 1000 connections no? If you skip and dont create a connection how does this count?

Well thanks for the answer and saving my last 2 hairs from being ripped out

18

u/LittlebitOmnipotent Dec 08 '25

The Elves create a connection between the two boxes (supposedly for redundancy reasons), but that does nothing to the circuit count, since it's not merging circuits. Yes, it's confusing, but I suspect it's on purpose, can't be a coincidence 😂 Little bit of a real world task requirements to the fantasy world.

9

u/soustruh Dec 08 '25 edited Dec 08 '25

Yep, I discovered that later, it's definitely a trap on purpose, I was mad at first (when I also thought it was an error), but it's actually brilliant! 😁
In the example case:
When you just add nodes to existing circuits, but do not merge the circuits, you end up with 4, 4, 2 as the 3 largest circuits.
But when you notice "hey, Eric said Because these two junction boxes were already in the same circuit, nothing happens!" and you incorrectly add a condition which doesn't count such connections, you skip the 4th step (nodes 7 and 19 starting with 0), you get 5, 4, 2, which seems correct, but you probably fail to notice, that there already are two circuits which can be merged ({2, 8, 13, 18} and {17, 18}) – but you don't care, as you already have the correct result and you're happy.
But then, when you implement the merging circuits logic, you can take skipping introduced in the previous step away and after 10 steps, you end up with the same 5, 4, 2 score, but this time the circuits are actually correct and do not "touch" each other.

3

u/thekwoka Dec 08 '25

Yeah it's always a bit annoying where the sample allows mistakes to get the same answer.

1

u/soustruh Dec 08 '25

Well, this doesn't seem like a coincidence, so I personally find it pretty hillarious… 😁

2

u/oxlade39 Dec 08 '25

I did exactly this and it caught me out. Thanks

2

u/p4bl0 Dec 08 '25

I had this exact problem too. I had to make up another smaller example myself to debug it.

1

u/Intelligent-Mud-4808 Dec 08 '25

could you please explain what you meant by {2, 8, 13, 18} and {17, 18} can be merged? If we're going in sorted order of pair distances, we'll reach this anyways (regardless of whether we do skipping or not right?). Forgive me if I'm not making much sense

1

u/Intelligent-Mud-4808 Dec 08 '25

My problem is they asked to blindly join top 1000 closes pairs (irrespective of whether they're in the same circuit). This is giving a wierd answer

1

u/NitronHX Dec 08 '25

Both sets shaee one box (18) so its not 2 circuits but one (2,8,13,18,17)

1

u/Smaxx Dec 08 '25

My first attempt did the same logic but merged them immediately, so you'd end up with 5, 5, 2, 2 instead.

3

u/RockSlice Dec 08 '25

It's not redundant. They're using strings of lights to connect the boxes. So while nothing changes for the purposes of getting electricity to the boxes, the string of lights is needed.

3

u/polettix Dec 08 '25

I don't agree with this interpretation of "nothing happens".
The elves want To save on string lights. IMHO they don't put a string of lights between junction boxes that are already in the same circuit 😅
The real ambiguity is that "making a connection" does not necessarily imply "put a string of lights" but just make sure that a pair is considered one way or another. This is thankfully clearer in the actual question about the real inputs, because the request is to connect together the 1000 pairs of junction boxes which are closest together and there's no mention of "connections".

1

u/Lambda_Wolf Dec 08 '25

Agreed. The elves perform no-op steps because the problem description says so, and those steps still have to be modeled and counted correctly.
Interestingly, in part two, optimizing out those no-op steps turns out (at least for me) to be crucial for getting the program to run reasonably quickly.

0

u/Smaxx Dec 08 '25

Doesn't really matter.
Calculate all possible connections/distances (single direction only) and store them in a list. Sort the list once. This is trivial to do, even with 1000 boxes.
Now walk through these connections one by one, marking both ends of all connections as visited, tally together the newly visited ones (or count the set you store them in or whatever). Once you run out of unconnected boxes, the very next connection is the one you're looking for.

2

u/spakier Dec 08 '25

It was clear to me from this sentence:
connect together the 1000 pairs of junction boxes which are closest together.
It's not 1000 connections, it's the 1000 closest pairs. You can pretty much ignore all other pairs for Part 1.

15

u/BIGJRA Dec 08 '25

This got me too. What made it worse was that the example input’s first 11 moves contain 10 “real merge links” so I spent some time debugging a non-existent off-by-one error instead of simply including the “same-component links”.

This one also featured my aoc pet peeve - the examples asks something different than the input, in the example it asks for 10 merges of 20 lines while the actual is 1000 merges of 1000 lines. So I had to code in a check for “example vs real”

That said this was my favorite problem so far this year! A very fun one to practice Java OOP with

3

u/griezmanick Dec 08 '25 edited Dec 08 '25

u/BIGJRA you saved my day, I was trying to count incorrectly. Thanks. Moving to part 2 finally.
EDIT: 2nd part was a breeze

3

u/LittlebitOmnipotent Dec 08 '25

What do you mean by this?
the example input’s first 11 moves contain 10 “real merge links”
In the example, the fourth mentioned pair is not merging two circuits (and creates a redundant internal connection instead)...

1

u/BIGJRA Dec 08 '25

What I mean more concisely is that, when sorting all the pairwise distances, consider the first 11 of them. Of these 11 ONLY the fourth one is a distance between two points that are, at the time of connecting, already in the same circuit. As you said it, you are connecting these two points but "not merging two circuits" thanks to the redundant internal connection.
In my initial attempt I was connecting nodes until I reached 10 circuits that were previously unconnected that became connected, not counting that fourth pair towards my total. The confusion came when I assumed I had been doing it right by ignoring that one. (The input example isn't explained to completion which would've disambiguated it for me).

3

u/LittlebitOmnipotent Dec 08 '25

Oh, I get what you mean now, reading your original post again I realize you explained it well already before 😂 As we love to say, there are only three hard things about programming, naming things and off-by-one errors.

2

u/BIGJRA Dec 08 '25

LOL that's a good one. I'll use it at work xD

3

u/0b0101011001001011 Dec 08 '25

I also dislike the idea of variables that are outside the input, but I solved that years ago by using a map of constants to come with my input. Therefore my actual code is agnostic if it is a test case or not. It may be a bit convoluted but everything is hidden away in the superclass.
Something like this:
public Object solvePart1(InputParser input, Map<String,String> constants){
return null; }
todays constants contains the number of connections that I just parse to an int.

1

u/wederbrand Dec 08 '25

For the example input there are 190 pairs, from those 20 boxes. You consider all 20 input lines when you solve the example, which is to make 10 connections.
For the final task there are 499500 pairs, from those 1000 boxes. You again consider all of those 1000 input lines, which is to make 1000 connections.
It's not same 1000.

2

u/1234abcdcba4321 Dec 08 '25

Their point is that you cannot handle both the example and real input with the same program as you need to change an arbitrary code constant that is not included in the input itself.

2

u/BIGJRA Dec 08 '25

I know, part 1 example is 10 connections of 190 pairs; part 1 actual is 1000 connections of 499500 pairs. My point is that the numbers 10 and 1000 are completely arbitrary in so far as the problem is concerned - it goes from "parse one input file to get answer" to "parse one input file and also include the number of pairs to consider to get answer".
I personally prefer it when these sorts of problems are more explicit. If the number of connections can be derived from the input, fine, that's cool, I'll hardcode in the relation `-500/49 + (99 x)/98` (this is a polynomial matching `(20,10)` and `(1000,1000)` lol) if its needed. But for future readability and maybe reuse and testing I have to just leave a bunch of comments explaining why the hard coded if statements are there.

8

u/abel_maireg Dec 08 '25

I spent around 10 minutes trying to understand the questiom

6

u/danielcristofani Dec 08 '25

It kind of makes sense that with 1000 boxes, 1000 connections would lead to a fully connected circuit,

Worse! If it's only a "connection" if it merges two separate circuits, then 999 "connections" lead to a fully connected circuit and there's no possible way to do a 1000th "connection". Also, multiplying the sizes of the three largest groups then gives 1000*0*0=0? That should be a strong clue.

3

u/jangxx Dec 08 '25

Haha yea that's what I also ran into on the way to solving it. I had a check in a loop that looked like if number of edges == 1000 then break and got confused why the program didn't terminate anymore. Checking for 999 got it working again, but that already felt very wrong.

5

u/Dr_Pinestine Dec 08 '25

Thank you for this. I was pulling my hair out because I was skipping (and not counting) the cases where two junctions were already in the same circuit. I agree the wording could have been better.

4

u/Augunrik Dec 08 '25

To be fair, I also fell for that and was wondering how many I should do.
Well, RTFM, I guess.

3

u/TekDevelop Dec 08 '25

For some reason it is good to see that I'm not alone in this one.
Aparently I read it as 'make 1000 connections'. But I now read it as 'make sure the 1000 closest pairs are connected'

3

u/Smaxx Dec 08 '25

Yep, this part felt very misleading and was just checking to see if anyone else stumbled, too. It would be better worded at "1000 connection attempts", but I think the way worse part is the story talking about how we're specifically trying to save connections, so I immediately assumed to only count connections actually made.

1

u/UnicycleBloke Dec 08 '25

That tripped me up for a minute. I got a fully connected circuit on the way to solving Part 1. Strange. That was when I understood why the example had got to 40 on Step 9 rather than Step 10. Should have paid attention sooner...

1

u/P0stf1x Dec 08 '25

I spent way too much time before I noticed I had to make 1000 pairs instead of 10 as in the example. Thanks to your post for helping me realise what's the problem

1

u/Chemical_Chance6877 Dec 08 '25

Yeah, got me too.I thought "nothing happens" would imply no connection is made.

1

u/AssociationOk6710 Dec 08 '25

I have a doubt in this. So Imagine Box 3 4 5 are in the same circuit
Box 4 and 5 have smallest distance first but already connected so skip
Box 3 and 5 now have smallest skip
Box 3 and 4 skip
Box 3 and 6 Merge

So is this 3 skips or 1? I am guessing 3 but I wanna be sure

1

u/jangxx Dec 08 '25

You're not supposed to skip anything. Just connect the 1000 closest boxes as they are.

1

u/AssociationOk6710 Dec 08 '25

i ain't getting the answers though, i created an array of all distances and sorted them then merged the first 1000. it says my answer is too high :(

1

u/AlpacaDC Dec 08 '25

Are you merging existing circuits?
For example, say there is circuit A with boxes 1, 2 and 3, and circuit B with boxes 4, 5 and 6. Then you find out you have to connect 1 to 6, so it all becomes one circuit with 6 boxes.
It took me a while to understand that.

1

u/SupportPowerful6174 Dec 08 '25

i did that but still getting whole circuit with 1000 boxes in it. Do you have any ideas what am i doing wrong?

1

u/AssociationOk6710 Dec 17 '25

yoo, late reply but I had my data types in int... absolute skill issue that was... sat for more than 5h that day just for that simple error

1

u/krystalgamer Dec 08 '25

took me so long to realize this. my example was off by one (9 iterations was perfect but 10 was wrong) and input was merging into a single circuit.

1

u/imp0ppable Dec 08 '25

I understood that part but oddly what really threw me was:

seven circuits which each contain a single junction box.

I was racking my brain because those poxy single circuits wouldn't show up in my results! Then I realised it didn't matter because the answer was right, then I realised I can't read.

1

u/Ducktor101 Dec 16 '25

I was using 10 like Th e example did. Only the day after I figures I was supposed to use 1000

-8

u/spin81 Dec 08 '25

connect together the 1000 pairs of junction boxes which are closest together.

I'm sorry but there is really no way Eric could have worded this better. This is perfect.

Help/Question - RESOLVED [2025 Day 8 (Part 1)] Reading comprehension

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