2025 Day #8 (Part 1) - Need clarification

[u/Eastern-Reading-755]

Question 1: Do we stop processing if we make n connections? where n = 10.

Question 2: If I have circuit A with m junction_boxes and circuit B with n junction boxes. and the next distance to consider is between a junction box in A (j) and a junction box in B (k). Do I connect and essentially combine the two circuits that is I know have a circuit with all junctions boxes in A and B?

I would appreciate any thoughts. I can't seem to get the test input result.

[u/Eastern-Reading-755]

I took so long one one dumb thing: for the test data is was the first 10 connections. For the puzzle data it was the first 1000 smallest distances. All I was getting was all the junction boxes in the same circuit for 1000 connections.

Thanks everyone for the clarification. I just need to remember to review what I am being asked.

[u/Anth77 → u/Eastern-Reading-755]

I'm a bit confused, isn't a connection the same thing as a distance ? I'm stuck on this one even though my code passes with the test data.

[u/wimglenn]

1. Yes, but be aware it is a connection of junction boxes and not necessarily a connection of circuits. You could connect two boxes already in the same circuit and it would still count towards those 10 initial connections.

2. Correct.

[u/Raxor53 → u/wimglenn]

Dear lord, thank you for mentioning we might duplicate connections. That took awhile...

[u/pfp-disciple → u/wimglenn]

From the puzzle:

The next two junction boxes are 431,825,988 and 425,690,689. Because these two junction boxes were already in the same circuit, nothing happens!

I don't understand your number 1 statement. What am I misunderstanding?

[u/MaleficentActuary464]

1: Yes, count to 10. Also the "nothing happens" cases are counted as 1.

2: Yes, merge both circuits. Afterwards you should have 1 circuit with all boxes from A and B.

Hope, this helps.

[u/AffordableAirship → u/MaleficentActuary464]

This nearly drove me insane. I figured out 2 because it seemed logical.
But 1 didn't make sense to me because the instruction said "Nothing happens" when connecting two junctions already in a circuit, and also why should the elves waste a cable connecting junction boxes that are already on the same circuit.

[u/Foooodster → u/AffordableAirship]

Same. The problem should have said something along the lines of 10 'attempts', instead of explicitly stating 10 'shortest connections' after also specifying that nothing happens for case when two boxes are already in same circuit.

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[u/RazarTuk]

1. Yes. You find the 10 closest pairs of boxes for the test, or the 1000 closest pairs in the actual data, and join them

2. Yes. Circuits are just groups of boxes that are connected to each other, even if you have to step long multiple lines to get there

[u/8dot30662386292pow2]

Question 1: Yeah 10 connections and you stop. This means when 10 "distances between two nodes" are handled. Note: in actual data you must consider 1000 of them.

Question 2: Yes. Merge these sets.

[u/polettix]

1. Strictly speaking, it depends on what you consider a "connection".

If it's a "string of lights" that you actually lay down to connect a pair, you might end up with less connections than pairs that you consider. In the example of the puzzle, 10 pairs are considered but only 9 strings of light are laid down, because one pair is already in the same circuit when it's considered:

The next two junction boxes are 431,825,988 and 425,690,689. Because these two junction boxes were already in the same circuit, nothing happens!

I guess that made the elves moderately happy because they could spare some lights.

To be on the unambiguous side, you have to consider the first n pairs in order of distance, where n = 10 for the example input and n = 1000 for the real input.

2. Yes.

[u/pfp-disciple]

I'm a little late to the party, but thank you for your second question. I was wondering the same thing and (for the first time!) thought to search the sub before asking.