this post was submitted on 10 Dec 2024
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[–] [email protected] 2 points 5 days ago* (last edited 5 days ago) (3 children)

Day 13, day 13 of shirking other responsibilities.

p1Ok. So, I overthought this a little. Ultimately, this problem boils down to solving a system of 2 linear equations aka inverting a matrix.

Of course, anyone who has done undergraduate linear algebra knows to look to the determinant in case some shit goes down. For the general problem space, a zero determinant means that one equation is just a multiple of the other. A solution could still exist in this case. Consider:

a => x+1, y+1
b => x+2, y+2
x = 4, y = 4 (answer: 2 tokens)

The following has no solution:

a => x+2, y+2
b => x+4, y+4
x = 3, y = 3

I thought of all this, and instead of coding the solution, I just checked if any such cases were in my input. There weren't, and I was home free.

p2No real changes to the solution to p1 aside from the new targets. I wasn't sure if 64 bit ints were big enough to fit the numbers so I changed my code to use big ints.

I'm looking at my code again and I'm pretty sure that was all unnecessary.

[–] [email protected] 3 points 4 days ago* (last edited 4 days ago) (1 children)

I liked day 13, a bit easy but in the right way.

Edit:

SpoilersAlthough saying "minimum" was a bit evil when all of the systems had exactly 1 solution (not necessarily in ℕ^2), I wonder if it's puzzle trickiness, anti-LLM (and unfortunate non comp-sci souls) trickiness or if the puzzle was maybe scaled down from a version where there are more solutions.

[–] [email protected] 2 points 4 days ago

spoilerGiven the lack of edge cases, I feel the latter possibility is strong. I'm just glad it was easy!

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