We're asked to insert numbers into a circular list, skipping a certain amount of entries each time (the skip number might be smaller than the length of the list). For part 1, we insert 2018 values and find the number right after the last one inserted. For part 2, we insert 50,000,000 values and find the number right after 0.
Part 1 was easily done with a basic python list and modulo math to simulate it being circle. The number of values was small enough that choice of data structure didn't matter.
Part 2 would not work after changing the Part 1 code from 2000 to 50 million. Inserting at a arbitrary list index in python is O(n), which is too slow for inserting 50 million at various parts in the list.
First fix: Change list() to blist(), a more efficient list that offers O(log n) insert performance. Without any changes, part 2 ran in 45 seconds. This was acceptable, but I wanted to find a better solution.
Second fix: Upon reflection, we don't need the vast majority of the numbers in the list to answer the question. So why bother inserting them? I changed the program to loop up to 50 million without actually building the list; only the insert position and current number are tracked.
The problem asks for the number directly after 0. According to the rules of
the problem, 0 will always be the first entry in the list and never moves. So
the answer is always the entry at list[1]. I changed the loop to check if
it ever would insert in the 1 position, and if so, track that as the most
recent valid answer. When it reaches 50 million, it returns that as the
answer. Result: Program now finishes in 5 seconds and uses hardly any memory
at all.
2017 Day 17 on AdventOfCode.com
Suddenly, whirling in the distance, you notice what looks like a massive, pixelated hurricane: a deadly spinlock. This spinlock isn't just consuming computing power, but memory, too; vast, digital mountains are being ripped from the ground and consumed by the vortex.
If you don't move quickly, fixing that printer will be the least of your problems.
This spinlock's algorithm is simple but efficient, quickly consuming everything in its path. It starts with a circular buffer containing only the value 0, which it marks as the current position. It then steps forward through the circular buffer some number of steps (your puzzle input) before inserting the first new value, 1, after the value it stopped on. The inserted value becomes the current position. Then, it steps forward from there the same number of steps, and wherever it stops, inserts after it the second new value, 2, and uses that as the new current position again.
It repeats this process of stepping forward, inserting a new value, and using the location of the inserted value as the new current position a total of 2017 times, inserting 2017 as its final operation, and ending with a total of 2018 values (including 0) in the circular buffer.
For example, if the spinlock were to step 3 times per insert, the circular buffer would begin to evolve like this (using parentheses to mark the current position after each iteration of the algorithm):
- (0), the initial state before any insertions.
- 0 (1): the spinlock steps forward three times (0, 0, 0), and then inserts the first value, 1, after it. 1 becomes the current position.
- 0 (2) 1: the spinlock steps forward three times (0, 1, 0), and then inserts the second value, 2, after it. 2 becomes the current position.
- 0 2 (3) 1: the spinlock steps forward three times (1, 0, 2), and then inserts the third value, 3, after it. 3 becomes the current position.
And so on:
- 0 2 (4) 3 1
- 0 (5) 2 4 3 1
- 0 5 2 4 3 (6) 1
- 0 5 (7) 2 4 3 6 1
- 0 5 7 2 4 3 (8) 6 1
- 0 (9) 5 7 2 4 3 8 6 1
Eventually, after 2017 insertions, the section of the circular buffer near the last insertion looks like this:
1512 1134 151 (2017) 638 1513 851
Perhaps, if you can identify the value that will ultimately be after the last value written (2017), you can short-circuit the spinlock. In this example, that would be 638.
What is the value after 2017 in your completed circular buffer?
The spinlock does not short-circuit. Instead, it gets more angry. At least, you assume that's what happened; it's spinning significantly faster than it was a moment ago.
You have good news and bad news.
The good news is that you have improved calculations for how to stop the spinlock. They indicate that you actually need to identify the value after 0 in the current state of the circular buffer.
The bad news is that while you were determining this, the spinlock has just finished inserting its fifty millionth value (50000000).
What is the value after 0 the moment 50000000 is inserted?