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Advent of Code: 12/5/2023 18:57:06


DanWL 
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I didn’t do any optimisation. Made something to better track how long it takes for the solutions to run. I felt like by the time it took to implement the optimisation, the program would have ended by then.
Advent of Code: 12/5/2023 21:14:27


DanWL 
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I only save TDB from making sure seed ranges are unique. Would be 15,845,578,766 total seeds and that took 48 mins 22 secs 842 ms (with only Notepad++ and command prompt running). Is saving of TBD% (rounded to 2 dp).

Edit: there was a bug.

Edited 12/5/2023 21:24:50
Advent of Code: 12/6/2023 05:44:46

JustinR17 
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wow the difference between day 5 and 6 😂 I'm surprised they made the input so short and brute-forceable
Advent of Code: 12/6/2023 19:00:43


DanWL 
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I thought day 6 could be so easy I could do it while bevvied. Is just 2 arrays, times and differences. For each possible time, compare how long is possible to go.
Advent of Code: 12/6/2023 19:04:04

JustinR17 
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Spoilers below for anyone that hasn't done it yet:

Or you can break out the quadratic formula for a more direct solution without iteration :)

Edited 12/6/2023 19:04:16
Advent of Code: 12/6/2023 21:31:11


Just_A_Dutchman_ 
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I am already falling behind, was bug fixing till 4:00 AM for day 3, second problem...

I am slowly regretting using C... lmao

Edited 12/6/2023 21:32:28
Advent of Code: 12/7/2023 02:06:05


DanWL 
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i think the real challenge of day 6 is to halve the number of iterations. It's a bell curve.
Advent of Code: 12/8/2023 13:02:55


DanWL 
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Is day 8 part 2 another of these CPU intensive tasks? Before was just day 5 part 2.
Advent of Code: 12/8/2023 14:51:50


FiveSmith 
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Yes, in the sense that the computation complexity serves as a restriction to motivate participants to make a smarter algorithm for the task.
No, in the sense that it cannot be brute forced, as "easily" as day 5.
Advent of Code: 12/8/2023 15:00:41


DanWL 
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I don’t know another way to solve it. My program has been running for about 2 hours.

Uses a while loop which contains a for loop to follow the complete path from the starts to an end.
Advent of Code: 12/8/2023 15:18:26


FiveSmith 
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You are expected to notice, that each of the ghosts starts visiting the same (for him) exit node again and again after certain amount of steps, i.e. that he is wandering in loops. The task is to identify those loops for each of the ghosts, and calculate the step count, where all the loops simultaneously end on the exit nodes.
Advent of Code: 12/8/2023 15:26:44


DanWL 
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Do you only need to find out where each ghost’s path end and do a calculation based on that?
Advent of Code: 12/8/2023 15:38:07


FiveSmith 
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The paths of each of the ghost are infinite. You need to identify loops for each of the ghosts:
- exit node
- step, when the ghost lands on the exit node
- loop size: steps between each revisit of the end node

Having those data on each of the loops allows for a quick calculation, on when all loops will converge
Advent of Code: 12/8/2023 15:56:30


DanWL 
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Based on the output I got after the optimisation, the number is way to high to go through one at a time. It's more likely for my laptop to randomly crash before getting the answer.

I left GCSE maths buried in the past.

My answer is in the 10s of trillions.

Everything up to this part I didnt need google for.

Edited 12/8/2023 16:10:04
Advent of Code: 12/9/2023 01:16:25


l4v.r0v 
Level 59
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Anyone doing Hanukkah of Data/Noah's Rug this year?
Advent of Code: 12/9/2023 02:13:37


DanWL 
Level 63
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Never heard of those.
Advent of Code: 12/9/2023 15:56:18


FiveSmith 
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Day 9 was disappointingly easy. Or maybe I got too lucky, that my straight implementation didnt encounter any hidden issues.

Edited 12/9/2023 15:57:31
Advent of Code: 12/9/2023 17:29:07


DanWL 
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Multiple levels of arrays are confusing.
Advent of Code: 12/16/2023 14:34:25


FiveSmith 
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Has the WZ AoC interest ran out of steam? Besides me, there is barely any star-progression during the past week.
Advent of Code: 12/16/2023 14:36:54


DanWL 
Level 63
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I’m away for until sometime next week. Can’t exactly code on mobile.
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