Advent of Code 2023 Day 24 (BaseA Style)

Discussion thread for day 24 of the Advent of Code - https://adventofcode.com/2023/day/24

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Part 1 is fairly straightforward vector intercept calculation (recommend converting to slope/intercept form and then comparing.)

Part 2 is a linear algebra mess that is probably better solved in Mathematica than Alteryx.

One hint for part 2:

Spoiler

Remember how many points it takes to define a line in 3d space. Pick convenient ones and chuck the rest.

Qiu

Agree. Part 1 is just about a formula about the two lines intersecting point.
For part 2, unfortunetly, Linear Algebra is not my thing 🤣....

Gawa

I did MATH, rather than building WF. No macro solution.

Part1

Spoiler

Just solve linear algebra. Not a big deal.

Part2

Spoiler

1) Find the pair of ices having the same velocity component for x, y, z for each
*following explanation will be based on x coordinate, but do the same thing for y and z, too.

2) Calculate the distance between the pair(px1-px2). That distance shall be divisible by (rock_vx)-(vx)
Refer to the below image for better understanding why we can say so.

3) Find the rock_vx by "Brute Force". For all of possible rock_vx(say 1-1,000), I checked if Mod([px1]-[px2],[rock_vx]-[vx])=0 ,and find the rock_vx having this value is 0 for all of pairs, that will be rock_vx.(In othere word, your task is to find common divisor for all of pairs)

4) Repeat this process for y, and z, then now you have (rock_vx, rocl_vy, rock_vz). Next, we want to know the (rock_px, rock_py, rock_pz), that is the answer for part2.

From here, it might be not the case for you, but my dataset worked this way

5) In my dataset, I fortunately found the ice having vy=rock_vy (or, maybe deliberately by AoC). We know that the rock hit to this ice, so, we can immediately conclude that rock_py shall be py of this ice, otherwise rock and this ice will never hit in the future.
*Imagine, two cars are running straight at the same speed, and toward the same direction in parallel. If their positon are different(one ahead/behind to another), two cars never hit each other, even if one car changes its lane to other lane where another car is running. Only when their position is the same, two cars will hit when trying to change the lane.

6) We have rock_py, so pick up any ice except for one we used in previous step, then do the MATH like; found time T when that ice will hit, and other unknown variables, including rock_px and rock_pz, will be determined automatically.

Overall WF

Spoiler

estherb47

Not a pleasant one, and I can't figure out what's wrong with my WF (Answer is too high)

Anyone care to help? @gawa , @Qiu , @ScottLewis ? Workflow attached

I've had quite a few "edge case" type data sets, though I can't see how that would be impacting my calculations in a straightforward-ish linear algebra problem....

Thank you!!

AOC24.yxmd

ScottLewis

@estherb47

You're appending a fileset to itself to create a full match. This works, but it gives you two rows for every pair of input rows (A->B & B->A.) If you don't account for that you're going to double count everything (which you are, checked your workflow with my dataset.)

The way I like to correct for this is to do a record ID before the append and then filter on LeftID>RightID. That also removes the self matches (and can be changed to >= if you want to include them.)

estherb47

@ScottLewis , thank you! I was wondering if that were the problem. Thanks for confirming. I usually like to solve for that with a Generate Rows and then join on the record IDs

Qiu

@estherb47
I agree with @ScottLewis and that is how your result is doubled. 🤣

I usually do is to create another parameter by using Min and Max to pair A and B.
Now it gives correct answer for my input.
and I almost give up on Part 2.

2023-12-27 074310.png

AOC24-r1.yxmd

Gawa

@estherb47 This is a typical pitfall for me, that is simple one but cannot find it immediately😅

As @Qiu and @ScottLewis mentioned, I always go with "Appendix Field + Filter" method to create the unique pairs without duplication.

Good luck!

CoG

This problem killed me XD I built a Big Num system in Alteryx which handles Multiplication, Division, and Addition (Subtraction is just addition with sign change).

Wrote out the formulae to generate a system of equations, then used Alteryx to calculate the solution to the system of equations for single variable. The equations are the same whether solving for x,y, or z so substitute relevant variables and the answer is done. One of the worst possible ways to solve, but I did it, and I'm quite happy about that.

Also noticed a cool graph pattern that I couldn't think how to exploit:

Spoiler

Each graph of position vs velocity forms a sort of hour glass shape, where the center point actually gives you the rocks starting position and velocity. I wasn't sure how to capitalize on this, but it is quite cool!
_X graph.png

Workflow:

Spoiler

Main Workflow:
_Main 1.png

Solution.yxzp

AkimasaKajitani

Part 1 is simple math problem. But Part 2 is not simple problem.

I made the mathematical equation for part 2, but the number is too big, so that complex formula didn't work. I wasted much time to do this.

Of course, I tried the fixed number.

Honesty, I couldn't solve this quiz without @gawa 's hint!

Spoiler

Part 1 : simple math

Part 2 :

In other programming languages, the solution would be Solver for simultaneous equations, etc., but Alteryx does not have that. So, we will use a trick (this trick does not work with the sample data because there is too little data).

Spoiler

I first considered some special cases. That is, there are several hailstorms with the same axial velocity. Let's find them first. Then, hailstones with the same axial velocity will always have a constant distance. And the distance will be a multiple of the velocity difference from the rock we throw (we use the MOD function for this).

MOD(([px]-[Source_px]),([SPEED]-[vx]))&#8203;

The rocks we throw are created using brute force (we are experimenting with a value of ±1000), and we find the speed at which all hailstones with the same speed meet this condition.

This gives us vx, vy, and vz for the rock.

Also, if the rock and the hailstone have the same velocity, they will only collide when they have the same initial point, so we can now find the initial value of one of the axes (in my data, the z axis matched).

Now we can solve the equation and we are done.

First, bring in the appropriate hailstone and find the collision time. This time, since I knew the initial value of z (pz), the equation to find the following t1 could be formulated by saying that the z-coordinates of the hailstone and the rock match.

t1: vz2, pz2 is the appropriate hail velocity, initial position

([pz2]-[pz])/([vz]-[vz2])&#8203;

Next, py. Since I knew the time, py is obtained by the following equation, where the position of the y-coordinate at time t1 coincides with that of the hailstone and rock.