General Discussions

AlteryxCommunityTeam · ‎12-11-2023

Discussion thread for day 12 of the Advent of Code - https://adventofcode.com/2023/day/12

gawa · ‎12-13-2023

Finally finished D12P2. As 2 days has already passed, I upload my WF this time.

My WF looks very simple, but needed uncountable attempts to make this....

DaisukeTsuchiya · ‎12-14-2023

@Pang_Hee_Choy - Thank you so much for sharing super straightforward your solution.

I was stuck for so long time then I can get idea from your spoiler although I didn't download your WF.
Finally I can finished after repeating debugging so many times. WF runs in 15 sec.
I can sleep tonight.

Spoiler

AkimasaKajitani · ‎12-16-2023

I struggled with Day12P2 but I finally got the second star!!!

I did not notice that the RegEx Match expression in a macro gave up to run, so it took me for long time.

Spoiler

Part 1 macro : simple macro

Part 2 macro:

In part 2 macro, I made it according to following term.

1. I convert rom ? to . and #
2. Create a definitive pattern on my own & delete the left-most number if it is already definitive.
3. Re-create Num.
4. Discard those that don't fit in the pattern check
5. Reduce the number of records by grouping and summing counts

This workflow takes 1 minutes to finish.

Part 1 macro : simple macro Part 2 macro: In part 2 macro, I made it according to following term.1. I convert rom ? to . and #2. Create a definitive pattern on my own & delete the left-most number if it is already definitive.3. Re-create Num.4. Discard those that don't fit in the pattern check5. Reduce the number of records by grouping and summing countsThis workflow takes 1 minutes to finish.

SeanAdams · ‎12-19-2023

This is one of those days where it started out with brute force, and then Part 2 showed me the error of my ways.

It’s now running part 2 in ~17 sec – very happy with this solution!

Spoiler

Solution:

A few things that will help you as you think through this:

Alteryx works with spaces much easier than . characters
A double . does the same job as a single – but it’s much easier to work with singles (because you can do a simple text match vs. target)
Leading & trailing . characters don’t add value but they do add complexity

So here’s the recipe:

Parse the data

change . to spaces;
remove duplicate & leading / trailing spaces
Build a target string of what you’re looking for – e.g. 1,1,3 will be #.#.###

Figure out valid combinations:

Repeat until you run out of strings with ? characters:

For each incoming row, Create 2 rows – one that replaces the first ? with a space; and one that replaces it with #
Trim leading & trailing spaces
Trim duplicate spaces
Throw away any strings which are shorter than the target
Figure out where the next ? is – call this X
Match the first X of my working string vs. the target – if it fails, discard
Then check if any of our working strings are the same – if so, then keep a counter
1. E.g. instead of having 2 rows with ###??? – instead keep 1 row with a counter of 2
Loop

This works fast for a few reasons:

Your number of rows that you are iterating with is very small because you compress them into a summary (step 7 above) and you are throwing stuff away as you go along
You are using simple string matches each time (all the complexity of leading and duplicate spaces is gone)

Solution: A few things that will help you as you think through this: Alteryx works with spaces much easier than . characters A double . does the same job as a single – but it’s much easier to work with singles (because you can do a simple text match vs. target) Leading & trailing . characters don’t add value but they do add complexity So here’s the recipe: Parse the data change . to spaces; remove duplicate & leading / trailing spaces Build a target string of what you’re looking for – e.g. 1,1,3 will be #.#.### Figure out valid combinations: Repeat until you run out of strings with ? characters: For each incoming row, Create 2 rows – one that replaces the first ? with a space; and one that replaces it with # Trim leading & trailing spaces Trim duplicate spaces Throw away any strings which are shorter than the target Figure out where the next ? is – call this X Match the first X of my working string vs. the target – if it fails, discard Then check if any of our working strings are the same – if so, then keep a counter E.g. instead of having 2 rows with ###??? – instead keep 1 row with a counter of 2 Loop This works fast for a few reasons: Your number of rows that you are iterating with is very small because you compress them into a summary (step 7 above) and you are throwing stuff away as you go along You are using simple string matches each time (all the complexity of leading and duplicate spaces is gone)