2023 Day 9 · advent of code · Zulip Chat Archive

Was pretty easy. My current solution basically does both Part 1 and 2 twice, one for each output, but it takes only a few seconds to run, so it's good enough. I start to like that you are not going to know how hard it will be.

timotree (Dec 11 2023 at 07:50):

Ryan Bates said:

You don’t need to calculate the difference of every value until they are all zero. Instead you can start at the last value and calculate the difference pyramid backwards until you hit a single zero.

Doesn't this break if there is a row ending in zero before the all-zeroes row? e.g.

timotree (Dec 11 2023 at 07:53):

My day 9: https://github.com/timotree3/aoc2023/blob/main/roc/day9.roc

Ryan Bates (Dec 11 2023 at 18:50):

Reply to timotree

Norbert Hajagos (Feb 08 2024 at 18:30):

I have found out something very cool about part2!
(It's been monts, so I'll not put this under a spoiler)
You don't need to first calculate the difference lists all the way until zeroes to start calculating the prediction per line "from the tip of the piramid back to the base of it". You just need to sum up the first numbers in the difference lists as they appear with alternating sign. So for the test input below, sum up the numbers in inside the stars (the first number is positive) :

 *10*  13  16  21  30  45
    *3*   3   5   9  15
      *0*   2   4   6
         *2*  2   2
           *0*   0

The prediction is +10-3+0-2+0 = 5.
Because the algorith for the i-th line prediction is

$x_0 = 0$ $x_i = - x_{i-1} + y_i$

where y_i is the i-th line's first number
You can find my solution here: https://github.com/HajagosNorbert/advent-of-code-2023/blob/main/day09.roc#L37-L44

Last updated: Jul 06 2025 at 12:14 UTC