2017-12-17 16:00:37 +00:00
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Advent of Code
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--- Day 17: Spinlock ---
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Suddenly, whirling in the distance, you notice what looks like a massive,
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pixelated hurricane: a deadly spinlock. This spinlock isn't just consuming
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computing power, but memory, too; vast, digital mountains are being ripped
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from the ground and consumed by the vortex.
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If you don't move quickly, fixing that printer will be the least of your
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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 the
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value 0, which it marks as the current position. It then steps forward
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through the circular buffer some number of steps (your puzzle input) before
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inserting the first new value, 1, after the value it stopped on. The inserted
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value becomes the current position. Then, it steps forward from there the
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same number of steps, and wherever it stops, inserts after it the second new
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value, 2, and uses that as the new current position again.
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It repeats this process of stepping forward, inserting a new value, and using
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the location of the inserted value as the new current position a total of
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2017 times, inserting 2017 as its final operation, and ending with a total of
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2018 values (including 0) in the circular buffer.
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For example, if the spinlock were to step 3 times per insert, the circular
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buffer would begin to evolve like this (using parentheses to mark the current
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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 then inserts
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the first value, 1, after it. 1 becomes the current position.
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* 0 (2) 1: the spinlock steps forward three times (0, 1, 0), and then
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inserts the second value, 2, after it. 2 becomes the current position.
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* 0 2 (3) 1: the spinlock steps forward three times (1, 0, 2), and then
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inserts the third value, 3, after it. 3 becomes 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 near
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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 the last
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value written (2017), you can short-circuit the spinlock. In this example,
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that would be 638.
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What is the value after 2017 in your completed circular buffer?
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2018-03-15 15:08:01 +00:00
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Your puzzle answer was 1642.
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2017-12-17 16:00:37 +00:00
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--- Part Two ---
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The spinlock does not short-circuit. Instead, it gets more angry. At least,
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you assume that's what happened; it's spinning significantly faster than it
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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 after 0
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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 just
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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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2018-03-15 15:08:01 +00:00
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Your puzzle answer was 301.
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2017-12-17 16:00:37 +00:00
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Both parts of this puzzle are complete! They provide two gold stars: **
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At this point, you should return to your advent calendar and try another
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puzzle.
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Your puzzle input was 301.
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References
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Visible links
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. http://adventofcode.com/
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. http://adventofcode.com/2017/about
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. http://adventofcode.com/2017/support
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. http://adventofcode.com/2017/events
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. http://adventofcode.com/2017/settings
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. http://adventofcode.com/2017/auth/logout
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. http://adventofcode.com/2017
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. http://adventofcode.com/2017
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. http://adventofcode.com/2017/leaderboard
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. http://adventofcode.com/2017/stats
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. http://adventofcode.com/2017/sponsors
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. http://adventofcode.com/2017/sponsors
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. https://en.wikipedia.org/wiki/Spinlock
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. http://adventofcode.com/2017
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