Advent of Code 2024 Day 14 (Base A Style)

Discussion thread for day 14 of the Advent of Code - https://adventofcode.com/2024/day/14

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Mixed feelings about this problem. Part 1: Another fantastic math problem - Part 2: Cool concept that was not executed well (especially based on what the actual answer is).

(NOT SPOILER): Immediately below is just a hint as to what to look for when doing Part 2:

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(SPOILER) Below is my solution to this problem with explanation for Part 1 and a short accounting of how incredibly lucky I got for Part 2:

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Part 1 was great! Good exercise in Modular Arithmetic. Because the robot "wraps around" whenever it arrives at a boundary in the map, its position becomes its out of bounds position mod the size of that dimension of the map (101 or 103 in the case of this problem). That alone is helpful to understand, but not what makes this problem a breeze to solve. The key insight is that modulus is a periodic function and thus it is irrelevant when you move the robot. We can thus calculate all 100 steps in one Formula Tool ([Position] + 100 * [Velocity]) and then take the modulus to determine the robots ultimate position at the end of the 100 seconds. One caveat is that in Alteryx the mod() function can output negative values, which must be accounted for, and can be by performing a double modulus:
Mod( Mod( [Position] , [MapSize] ) + [MapSize], [MapSize])

The inner Mod() may output a negative number, and so by adding the modulus ([MapSize] in this case) we can correct for that guaranteeing that the output is positive. The second Mod() is needed to account for the overshooting that occurs when the inner Mod() was positive.

From there, we calculate the quadrant, then the product of quadrant counts for the final answer!

Part 2: Not knowing what the Christmas Tree was supposed to look like, I thought the entire map might be filled with all robots making a centered Christmas tree, implying that at least one would be at position (0,50). Second, it is true that all robots will be on their starting squares after 101*103 = 10403 seconds (so the answer had to be within this range), so I used Generate Rows Tool to calculate 9000 single second time stamps, then Filtered down to only those cases where that position was occupied, which left me with 426 timestamps to check based on my initial premise within that range. That position was 100% irrelevant to the Christmas tree, but by an act of pure providence, the answer happened to have a robot sitting at (0,50), and I found the tree. The timestamp that I found due to the periodicity of the problem, meant that I had an upper bound on what the solution could be, but not wanting to be bothered to verify, I submitted my answer, and it happened to be right. So... victory, I guess....

Main.yxmd

clmc9601

Shocking to suddenly see a Christmas tree in my results. 😂 I'm very pleased to have figured out the pattern!

Spoiler

Part 1: thankfully, my AoC experience paid off and I knew I could solve this with modulo instead of building an iterative macro. Took some adjusting to get the wrapping around the edge of the board right.

Part 2: I started out by visualizing the first 600 steps to find the pattern. I use the X and Y coordinates as spatial lat/long integers. I prefer using the Report Map tool more than the charting tools.

Want help finding the pattern? I'll include an explanation below.

Spoiler

The trick is to find the intersection of the X wraparound line and the Y wraparound line.

After looking at the first 600 steps to find the pattern, I found that my lines (y = mx + b) were as follows:
X wraparound: width of board * number of seconds + 48
Y wraparound: height of board * number of seconds + 1

I then generated the first few multiples of these lines to find the intersection.

Surprise! Suddenly a Christmas tree

2024_AoC_clmc9601_Day_14.yxmd

ScottLewis

Part 2 is the worst Advent puzzle I have ever seen. Not the hardest, the worst. They may be going a bit harder than is reasonable on "problem cannot be solved by AI."

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If you're having trouble finding the tree, try looking for horizontal lines. A line of 5+ robots should be rare in your data. I charted the 5+ line steps and looked for the tree but you could probably just look for a line of 7-10 and get it down to a single case.

AkimasaKajitani

Solved!

My Viz

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I think part 2 is set as an easter egg so the condition is intended.

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My workflow for part 2 is intentionally a bit blurry.

Answer

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All robot positions should be unique. Because it is an Easter Egg?

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Gawa

I sense too strong smell of 'anti-AI' from part2. Thanks to @AkimasaKajitani for hints.

Spoiler

As 101 and 103 are relatively prime, positioning will repeat after every 101 x 103 seconds.
Find the timing when all robots locates at distinct positions(different 500 positions), and check how they look like. In my data, there were two.

(EDIT)
How to somewhat quantitatively derive answers without relying on visual aid.

Spoiler

I focused on proximity of robots when they are arranged in Christmas tree's shape.
Sum up the distance to the nearest robot for all 500 robots, and plot it by elapsed times.(Create spatial objects and use the Find Nearest tool)
Shorter distance means robots come closer and longer distance does robots locate far each other.
Obviously there exist a single minimum point that implies they come closer to make Christmas tree.

2024D14_gawa.yxmd

DaisukeTsuchiya

Today's AoC was interesting.

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P1 can be solved if think carefully without rushing. However, I rushed through it, so it took me more time than I expected.

For P2, I used Map Tool and Render Tool to create a PDF for each loop and checked them. However, after outputting 5000 loops and reviewing just a few hundred pages, my hand started cramping. During this process, I noticed that vertical and horizontal stripes occasionally appeared. By analyzing their patterns, I discovered that these stripes appeared at intervals matching the number of tiles in the horizontal and vertical numbers.

The attached PDF contains only the extracted periods of these intervals. I was able to spot the Christmas tree after reviewing just 130 pages, without having to go through thousands of pages.

スクリーンショット 2024-12-14 170319.png スクリーンショット 2024-12-14 170352.png

DT2024_Day14_Rev1.yxmd

Extracted.pdf

Yoshiro_Fujimori

Today I learned another meaning of Easter Egg .

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Workflow
For Part2, I saw two vertical lines every 101 seconds.
So I added a filter to check only these images, and found a tree pattern in the 77th cycle of 101 seconds.

My image

workflow.yxmd

Goddenra

Enjoyed Part 1! Thanks to @DaisukeTsuchiya for the hint on the key to Part 2.

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dhavaldoshi

Spoiler

Part 2: Spoiler

I didn't use any reporting tools...

Assuming you arranged the characters to see the patter in Part 1, group by times, and then try to identify which iteration has a situation where there are most number of consecutive rows which have adjacent columns captured by a robot in acending or descending order, I got a Christmas Tree which was reversed. It's not fool proof as they could have purposely left multiple trees with one level missing, but this worked for me! Hope you find the easter egg too.

cgoodman

Spoiler

Part 1 I was able to approach it using mod() to deal with the wrapping and not need to generate every step to 100.
I used Excel to check that this was the right logic:
Screenshot 2024-12-14 123244.png

As @CoG mentions mod() in Alteryx can provide negative numbers, whereas Excel doesn't. To deal with negatives I used an if statement

if [newgrid_px] < 0 THEN 
[User.maxx] + [newgrid_px]
Else [newgrid_px]
ENDIf&#8203;

Part 2 is similar to day 10 in 2018, where there is a point where the stars align, or in today's case the robots.
After running a few 1,000 iterations I could see there were certain rows and columns which had a lot of robots in a line, so filtered to these. Then it was a case of increasing the iterations to cover the periodicy of the problem, which created a very clear outcome where the number of robots in line (I essentially was finding the frame around the Christmas tree) was the maximum.
Screenshot 2024-12-14 123633.png

Which gave me a nice solve without needing to hunt through 1,000s of images - so my workflow at the end creates a nice Christmas tree.
Screenshot 2024-12-14 123057.png

2024_AoC_ChrisGoodman_Day_14.yxmd

DataNath

Day 14 in the bag! Used @ScottLewis' tip for part 2 - looking for a line of 6 points returned a single output which was my Christmas tree! Not very neat so maybe my flow is off slightly but we picked up 2 stars!

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Hub119

Part 1 was pretty straightforward... Part 2 wasn't too bad once I figured out which logic to apply to it... and determined that it wasn't going to be ALL the robots that would align so shifted which metric I was tracking. 14 down, 11 to go!

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Workflow

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Tree Image

AoC 2024 Day 14.yxzp

mmontgomery

Day14. More in spoiler

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P1: Did a simple macro to account for each position since was 100 iterations
P2: Brute force 10k combos and write down two distinct patterns appearing horizontally and vertically, then figure out the delta between them and create possible combos with generate rows to narrow the search. I saw it visually in the report map

Day 14_MM.yxzp

DaisukeTsuchiya

I update WF for P2 without checking thousands loops.

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The task involved reassembling scattered robots, so I calculated the standard deviation of the XY coordinates for each loop and examined the smallest one. As expected, the case with the smallest standard deviation turned out to be the solution.

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DT2024_Day14_Rev2.yxmd

Tokimatsu

At first I could not imagine the goal picture.

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I tried to check for line symmetry placement, but failed. Next, they looked at 300 pictures in each sequence. Then we found a repeating pattern.Then found P2 answer. After that create the workflow to find the goal. スクリーンショット 2024-12-16 095114.png

AoC2024_Day14.yxzp

ccanterman

Solved. I'm not sure how I feel about Part 2. I wouldn't have gotten there without the help from everyone. I definitely would've preferred more guidance on what "make a Xmas tree" actually looked like, since I assumed it was set in the middle of the graph

Spoiler

For Part 2, I did a Generate Rows for 101 * 103 Rows. Then I did a Group By Iteration and End_X and Group By Iteration and End_Y. I filtered each for where the Count was > 15 (it seemed like a good number), joined them together, and found the first iteration on both.

Day 14.yxmd

kelsey_kincaid

This was challenging but fun!

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I spun my wheels for way too long on Part 1. I fixed the issue by simply adding 1 to all of my starting coordinates. Having zeroes was making the math confusing, and I wish I'd just adjusted the coordinates sooner. I managed the "teleports" by finding a multiplier that adjusted the coordinates.

Formula Examples:

For Part 2, I iterated 10,000 times and then found the iteration with the smallest standard deviation between points. I got a lot of inspiration for this solution and visualization from some previous year solutions from @phottovy and @cgoodman3 that they highlighted recently. Without that, I would have spun my wheels trying to figure this one out. Just goes to show how much of a learning opportunity Advent of Code is!

There's probably a way to get the macro to exit early instead of hardcoding the iterations, but this only ran in 25.6 seconds so I'm OK with it the way it is

2024_Day14_KelseyKincaid.yxzp

Qiu

@Tokimatsu
Very brilliant idea.
I did same approach but went to calculate the area of Christmas Tree.

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AOC2024-Day14B.yxzp

Shifty

Just part 1 for me. Part 2 seems mad.

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Image upload failed. Try uploading a different image or from some other source.
Image upload failed. Try uploading a different image or from some other source.
Image upload failed. Try uploading a different image or from some other source.

2024.14.yxmd

AncientPandaman

before see the picture, i totally have no idea about what the Christmas tree.
I thought is a words "CHRISTMAS TREE" like most of the time.

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part1: math

part2: count the adjacent cells of a robot that surround by another 4 robots (4 directions)
it had a huge vary between the 2nd place.
Screenshot 2024-12-18 154907.png

Day 14_Autosave.yxzp

Samantha_Jayne

Loving the number of spatial challenges this year.

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I found the tree, by looking at adjacency and finding the iteration with the most. It was a hunch that paid off.

Not sure why I am upside down though...

Day 14 - Both Stars.png

Samantha_Clifton_Day14.yxzp

estherb47

Past AOC challenges were definitely helpful with this one!

Spoiler

like many others, modulo was an easier calculation than point by point. For part 1, it was easy to calculate at the 100 second march with one single calculation

Part 2 was harder to conceptualize. Figured, eventually, that the tree would mean distinct points for each robot, so a point in the second with 500 unique robot positions. Having no idea where this would occur, decided to try width*height/2, and go within that range. Wasn't a large enough range, so went with 101*103 and found the correct point. Not the fastest or most elegant, but it works!

Day14.yxmd

Pilsner

I may have skipped a couple, but managed to get an answer for day 14.

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Macro

Day 14 part 1.yxmd

lwolfie

As I'm running out of time, I'll come back for the 2nd star.

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