93 lines
3.6 KiB
Markdown
93 lines
3.6 KiB
Markdown
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## \-\-- Day 17: Spinlock \-\--
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Suddenly, whirling in the distance, you notice what looks like a
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massive, [pixelated
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hurricane]{title="You know, as opposed to all those non-pixelated hurricanes you see on TV."}:
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a deadly [spinlock](https://en.wikipedia.org/wiki/Spinlock). This
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spinlock isn\'t just consuming computing power, but memory, too; vast,
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digital mountains are being ripped from the ground and consumed by the
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vortex.
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If you don\'t move quickly, fixing that printer will be the least of
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your problems.
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This spinlock\'s algorithm is simple but efficient, quickly consuming
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everything in its path. It starts with a circular buffer containing only
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the value `0`, which it marks as the *current position*. It then steps
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forward through the circular buffer some number of steps (your puzzle
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input) before inserting the first new value, `1`, after the value it
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stopped on. The inserted value becomes the *current position*. Then, it
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steps forward from there the same number of steps, and wherever it
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stops, inserts after it the second new value, `2`, and uses that as the
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new *current position* again.
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It repeats this process of *stepping forward*, *inserting a new value*,
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and *using the location of the inserted value as the new current
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position* a total of `2017` times, inserting `2017` as its final
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operation, and ending with a total of `2018` values (including `0`) in
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the circular buffer.
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For example, if the spinlock were to step `3` times per insert, the
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circular buffer would begin to evolve like this (using parentheses to
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mark the current position after each iteration of the algorithm):
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- `(0)`, the initial state before any insertions.
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- `0 (1)`: the spinlock steps forward three times (`0`, `0`, `0`), and
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then inserts the first value, `1`, after it. `1` becomes the current
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position.
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- `0 (2) 1`: the spinlock steps forward three times (`0`, `1`, `0`),
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and then inserts the second value, `2`, after it. `2` becomes the
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current position.
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- `0 2 (3) 1`: the spinlock steps forward three times (`1`, `0`,
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`2`), and then inserts the third value, `3`, after it. `3` becomes
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the current position.
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And so on:
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- `0 2 (4) 3 1`
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- `0 (5) 2 4 3 1`
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- `0 5 2 4 3 (6) 1`
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- `0 5 (7) 2 4 3 6 1`
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- `0 5 7 2 4 3 (8) 6 1`
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- `0 (9) 5 7 2 4 3 8 6 1`
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Eventually, after 2017 insertions, the section of the circular buffer
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near the last insertion looks like this:
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1512 1134 151 (2017) 638 1513 851
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Perhaps, if you can identify the value that will ultimately be *after*
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the last value written (`2017`), you can short-circuit the spinlock. In
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this example, that would be `638`.
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*What is the value after `2017`* in your completed circular buffer?
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Your puzzle answer was `725`.
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## \-\-- Part Two \-\-- {#part2}
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The spinlock does not short-circuit. Instead, it gets *more* angry. At
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least, you assume that\'s what happened; it\'s spinning significantly
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faster than it was a moment ago.
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You have good news and bad news.
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The good news is that you have improved calculations for how to stop the
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spinlock. They indicate that you actually need to identify *the value
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after `0`* in the current state of the circular buffer.
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The bad news is that while you were determining this, the spinlock has
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just finished inserting its fifty millionth value (`50000000`).
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*What is the value after `0`* the moment `50000000` is inserted?
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Your puzzle answer was `27361412`.
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Both parts of this puzzle are complete! They provide two gold stars:
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\*\*
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At this point, you should [return to your Advent calendar](/2017) and
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try another puzzle.
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Your puzzle input was `329`{.puzzle-input}.
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